A student has three springs A, B and C. He measures the length of each spring, in turn, when
different weights are placed on the end of each spring.
His table of results is shown in Fig. 3.1

The time required to accelerate this car from rest to a speed of 100 km/h at full power on a level road is 7.73 seconds.

Mass of the engine = 1500 kg

Power rating = 75 kW = 75,000 W

Final speed = 100 km/hr =  = 27.78 m/s

v₁ = 0

Power = Work done ÷ Time

Work done = Final energy - Initial energy

=1/2 x 1560 x 27.78^2 = 1/2 x 1560 x 0^2\

= 578703.72 J

Thus,

75,000 = 578703.70 ÷ time=7.72 seconds.

Work is done every time a force moves something over a distance. By multiplying the force by the distance traveled in the direction of the force, you can calculate the energy transferred or the work done. Energy transmitted = work completed = force x distance traveled in the direction of the force.

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The acceleration of the box is 6 m/s²; option A.

The acceleration of the box is determined as follows:

Frictional force = 660 * 0.15 = 99 N

Net force = 500 - 99 = 401 N

Mass of the box = 660/9.8 = 67.3 kg

Acceleration = net force/mass

Acceleration = 401/67.3

Acceleration = 6 m/s²

In conclusion, the acceleration of the box is determined from the net force and the mass of the box.

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Answer:

The resistance of the each bulb is 1.75 ohm.

Explanation:

The Power P supplied by the battery is given by the formula,

P=VI

where V is the voltage and I is the current.

Then the total current through given parallel circuit is I=P/V

Given P=48 W and V=6 V the value of I is

I=48 W/6 V

I= 8 A

Since both bulbs have same resistance and are connected in parallel so the current through each circuit will be half of the total current. Hence the current through each circuit I₀ is 4 A.

The Resistance R is calculated using formula R=V/I₀.

So the resistance is

R=6 V/4 A

R=1.75 ohm

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Answer:

The resistance of the bulb is 1.5 ohm.

Explanation:

The power P delivered to the whole circuit is given by the formula,

P=VI

Given  P= 48 W and V= 6V, the current in the circuit is

I=P/V

I=48/6= 8 A.

Since the bulbs are identical and are in parallel, therefore the current through each parallel circuit is half of the total current. Therefore current I1 through each bulb is 4 A.

The resistance is given by the formula,

R=V/I1

Hence R=6/4 = 1.5 ohm

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Let us assume the positive \hat{x} direction is towards the right.

During any collision, the total momentum is conserved.

That means, the total momentum before the collision is the same as the total momentum after the collision.

p before = P after

Given,

Mass 1, m_1 = 111.1kg

Mass 2, m_2 = 222.2kg

Initial Velocity of Mass 1, 1,1 3.00.î m/s

Initial Velocity of Mass 2, \overrightarrow v_{2,i} =0

Velocity of Mass 1 after collision, \overrightarrow v_{1,f} =-1.000\hat{x}\hspace{0.1cm}m/s

The total Momentum of the system before the collision is,

\overrightarrow p_{before} =m_1\overrightarrow v_{1,i} + m_2\overrightarrow v_{2,i}

=> \overrightarrow p_{before} =(111.1)(3.00\hat{x}) + 0

=> \overrightarrow p_{before} =333.3\hspace{0.1cm}\hat{x}\hspace{0.1cm}kgm/s

The total Momentum of the system after the collision is,

\overrightarrow p_{after} =m_1\overrightarrow v_{1,f} + m_2\overrightarrow v_{2,f}

=> \overrightarrow p_{after} =(111.1)(-1.000\hat{x}) + (222.2)\overrightarrow v_{2,f}

=> \overrightarrow p_{after} =(-111.1\hat{x}) + (222.2)\overrightarrow v_{2,f}

We know the total momentum before the collision is the same as the total momentum after the collision.

p before = P after

=>333.3\hat{x} =(-111.1\hat{x}) + (222.2)\overrightarrow v_{2,f}

=>(-111.1\hat{x}) + (222.2)\overrightarrow v_{2,f} = 333.3\hat{x}

=> (222.2)\overrightarrow v_{2,f} = 333.3\hat{x} + 111.1\hat{x}

=> (222.2)\overrightarrow v_{2,f} = 444.4\hat{x}

=> \overrightarrow v_{2,f} = \frac{444.4}{222.2}\hspace{0.1cm}\hat{x}\hspace{0.1cm}m/s

=> \overrightarrow v_{2,f} = 2.00\hspace{0.1cm}\hat{x}\hspace{0.1cm}m/s

Therefore the velocity of m_2 after collision is v_{2,f} = 2.00\hspace{0.1cm}m/s going to the right

Please check the attached file for a brief answer.

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With the use of the second equation of motion formula, the speed the car travelled when the rockets were fired is 12 m/s.

Linear motion depicts movement in a straight line. While speed is the distance travelled per time taken.

Given that In a test run, a rocket-powered car is driving down a test track at a constant speed when the rockets are fired. it then accelerates at 4 m/s2 for 8.0 s covering a distance of 224 m.

The given parameters are;

The speed the car travelled when the rockets were fired can be calculated by using the below formula

s = ut + 1/2a[tex]t^{2}[/tex]

Substitute all the parameters into the formula

224 = 8u + 1/2 x 4 x [tex]8^{2}[/tex]

224 =8u + 2 x 64

224 = 8u + 128

224 - 128 = 8u

8u = 96

u = 96/8

u = 12 m/s

Therefore, the speed the car travelled when the rockets were fired is 12 m/s.

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Answer:

1.8 = relative density (there are no units for relative density)

Explanation:

It displaces water equal to it's volume and gets buoyancy equal to that amount of water

2.7 - 1.2  = 1.5  N  of buoyancy

density of water = 1 gm /cc

 1.5 N = m (9.81)

    m of water displaced = .1529 kg

             152.9 cc of water will produce this buoyancy....this is the volume of the object

find mass of object     2.7 = m (9.81)     shows m = .2752 kg = 272.5 gm

density = mass/ volume = 272.5 / 152.9 = 1.8 gm / cc

Relative to water (which is 1 gm / cc) the relative density is 1.8

  ====> it is 1.8 times denser than water and will sink when in water....

Answer: A microwave oven operates at 2.4GHz with an intensity inside the oven of 2500 W/m^2. Then, the amplitudes of the oscillating electric and magnetic fields are, [tex]B_0=4.57*10^{-6} T and E_0=1.37*10^{3} V/m[/tex]

Explanation: To find the correct answer, we have to know more about the EM Waves.

                          [tex]E_0=cB_0[/tex]

                        [tex]I=\frac{1}{2u_0} cB_0^{2}[/tex]

                          [tex]B_0=\sqrt{\frac{2u_0I}{c} }=\sqrt{\frac{2*4*3.14*10^{-7}*2500 }{3*10^{8} } } =4.57*10^{-6} T[/tex]

                              [tex]E_0=cB_0=3*10^{8} *4.57*10^{-6} =1371 V/m[/tex]

Thus, from the above calculations, we get the amplitudes of the oscillating electric and magnetic fields are, [tex]B_0=4.57*10^{-6} T and E_0=1.37*10^{3} V/m[/tex]

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Answer:

você tem um camping de em Rio do Rio de Janeiro ou em lojas em SP e SP e RJ e região metropolitana de Porto Velho de Porto Velho Rio

Long distance running is the physical activity I engage in now, and there are a number of items and programs that can greatly improve my performance.

A GPS powered watch is one such item which I think is quite useful. I can monitor my speed, distance, heart rate and other important information while running with this device. This data lets me change my pace, track my effort, and make sure I stay within my desired heart rate range. The watch also provides post-run analysis, which I can use to identify my weaknesses and modify my training appropriately. In my opinion, GPS running watches have changed the game by allowing me to run harder and push my physical limits.

A strength training program designed for runners has also been very helpful in enhancing my performance. The program's primary emphasis is on functional workouts that target the major running muscle groups, such as the core, glutes and legs.

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Answer:

Weight is a body's relative mass, or the quantity of matter contained by it.

Explanation:

Hope this helps

Answer:

the gravitational force

Magnitude of force experienced by the 90Kg passenger is 450N.

To find the answer, we need to know about the pseudo force.

Pseudo force= 90×5=450N.

Thus, we can conclude that the force experienced by the passenger is 450N.

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So, the acceleration of the car is -1.25 m/s². In other word, the car is also decelerating by 1.25 m/s².

Hi ! I will help you to discuss about "deceleration in a straight line movement". Please note in advance that deceleration is acceleration which has a negative value. When an object decelerates, the object will continue to move until it reaches a certain speed (which is less than before) or until it stops. The higher the deceleration value, an object that is moving will stop faster and cover a shorter distance.

In this opportunity, I will give you the following equation to express the relationship between final velocity and initial velocity, acceleration, and distance.

[tex] \boxed{\sf{\bold{(v_t)^2= (v_0)^2 + 2 \times a \times s}}}[/tex]

With the following condition:

We know that:

Note :

What was asked ?

Step by step :

[tex] \sf{(v_t)^2 = (v_0)^2 + 2 \times a \times s} [/tex]

[tex] \sf{15^2 = 20^2 + 2 \times a \times 70} [/tex]

[tex] \sf{225 = 400 + 140 \times a} [/tex]

[tex] \sf{140 a = -175} [/tex]

[tex] \sf{a = \frac{-175}{140}} [/tex]

[tex] \boxed{\sf{\bold{a = -1.25 \: m/s^2}}} [/tex]

Here, we see that the acceleration is -1.25 m/s². In other words, the car is also decelerating by 1.25 m/s².

The Angle of Incidence is equal to the Angle of Reflection.

Take a wooden drawing board and fix a white sheet of paper on it. In the middle of paper draw a straight line and Mark a point B on it. Draw a perpendicular . Place a mirror on line such that the flat side of the mirror is along the line. Hold the mirror in the mirror holder.

Fix two steel pins P and Q on the straight line AB at least 10 cm apart. Look for the images of the pins P and Q and fix two pins P a such that P', Q', and images of P and Q are all in the same straight line. Remove the pins and draw small circles around the pinpricks.

Remove the mirror and Join P'Q' which produce the straight line to meet at B. Measure ∠ABN = i and ∠CBN = r. It is found that ∠i=∠r. This proves that the Angle of Incidence is equal to the Angle of Reflection.

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force is mass multiply by acceleration so it will be 150 multiply by 10 is 1500N

Answer:

Acceleration: [tex]100\; {\rm m\cdot s^{-2}}[/tex] assuming that the radius of the rotation is [tex]1\; {\rm m}[/tex].

Centripetal force: [tex]15\; {\rm N}[/tex].

Explanation:

In a circular motion, if the tangential velocity is [tex]v[/tex] and the radius of the motion is [tex]r[/tex], the centripetal acceleration of the motion would be [tex]a = (v^{2} / r)[/tex].

In this question, it is implied that for this circular motion, [tex]v = 10\; {\rm m\cdot s^{-1}}[/tex] while [tex]r = 1\; {\rm m}[/tex]. Thus, the (centripetal) acceleration would be:

[tex]\begin{aligned} a &= \frac{v^{2}}{r} \\ &= \frac{(10\; {\rm m\cdot s^{-2}})^{2}}{1\; {\rm m}} \\ &= 100\; {\rm m \cdot s^{-2}}\end{aligned}[/tex].

Note that the unit of mass in this question is gram, whereas the standard unit for mass should be [tex]{\rm kg}[/tex] (so as to leverage the fact that [tex]1\; {\rm N} = 1\; {\rm kg \cdot m \cdot s^{-2}}[/tex].) Apply unit conversion: [tex]m = 150\; {\rm g} = 0.150\; {\rm kg}[/tex].

Using that fact that [tex](\text{net force}) = (\text{mass}) \, (\text{acceleration})[/tex]:

[tex]\begin{aligned} (\text{net force}) &= (\text{mass}) \, (\text{acceleration}) \\ &= 0.150\; {\rm kg} \times 100\; {\rm m\cdot s^{-2}} \\ &= 15\; {\rm kg \cdot m \cdot s^{-2}} \\ &= 15\; {\rm N}\end{aligned}[/tex].

Answer:

The Average Velocity is 2.22m/s.

Explanation:

Average Velocity = displacement(m) ÷ time(s)

8km = 8000m

8km/hr = 2.22m/s

Finding Time

Time = distance÷speed

Time = 8000m ÷ 2.22m/s = 3603.6s = 3604s

Finding Average Velocity

Average Velocity = displacement ÷ time

Average Velocity = 8000m ÷ 3604s

Average Velocity = 2.22m/s

Therefore the Average Velocity is 2.22m/s.

The net force (magnitude and direction) of the system given in the question is 40 N horizontal to the right

Case 1 (Net force between up and downward force)

F1 = Fu - Fd

F1 = 50 - 30

F1 = 20 N up

Case 2 (Net force between right and left)

F2 = Fr - Fl

F2 = 60 - 20

F2 = 40 N right

SUMMARY

From the above, the net force between right and left (i.e 40 N) is greater than the net force between up and down (i.e 20 N)

Thus, the net force of the system will be 40 N horizontal to the right

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Answer:

-50N/m

Explanation:

Force , F = 15N

Displacement , x = 0.3m

Spring constant , K = ?

K = -F/x

K = -15N/0.3m

K = -50N/m

Answer: 50 N/m

Explanation:

Edge 2022

The time taken by the pulse to travel from one support to the other is 0.208 s.

The mass of the cord is m = 0.65 kg.

The distance between the supports is, d = 8.0 m.

The tension in the cord is T = 120 N.

The time taken by the pulse to travel from one support to the other is given as,

[tex]v=\frac{d}{t}[/tex]

[tex]t=\frac{d}{v}[/tex]

Here, v is the linear velocity of a pulse. Its value is,

[tex]v=\sqrt{\frac{T d }{m} }[/tex]

[tex]v=\sqrt{\frac{120 * 8}{0.65} }[/tex]

[tex]v= 38.43 m/s[/tex]

Then,

[tex]t=\frac{8}{38.43}[/tex]

[tex]t=0.208 s[/tex]

Thus, the time taken by the pulse to travel from one support to the other is 0.208 s.

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Answer:

time taken= 8s

deceleration = -0.625ms^-2

Explanation:

s = 1/2(u+v)t (linear motion formula)

20 = 1/2 ( 5+0 ) t

t=time taken, find t

t= 8s

a= (v-u) / t

a= (0-5) / 8s

= -0.625 ms^-2

10.7 rad/s is the final angular velocity of the stick.

Given:

Mass of the stick = 4.42 kg

Length of the stick = 1.23m

Force of impulse (I) = 12.8 N s

The linear velocity of the stick, [tex]v=\frac{I}{m}[/tex]

                                                  [tex]v=\frac{12.8 N.s (\frac{1 kg m/s^2}{1 N}) }{4.42 kg}[/tex]

                                                  [tex]v[/tex]  [tex]= 2.89 m/s[/tex]

Therefore, the final linear velocity of the stick is 2.89 m/s

∴[tex]w=\frac{12 Ir}{ml^{2} }[/tex]

[tex]w=\frac{12 ( 12.8 N.s ) ( 46.4 cm)}{(4.42 kg) (1.23 m)^2}[/tex]

[tex]w= \frac{12 (12.8 N.s) (46.4 cm) (\frac{10^-^2 m}{1 cm} )}{(4.42 kg) (1.23m)2}[/tex]

[tex]w=10.7 rad/s[/tex]

Therefore, 10.7 rad/s is the final angular velocity of the stick.

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Momentum is conserved when carts are collided on a slanting plane.

To find the answer, we need to know about the conversation of momentum.

Thus, we can conclude that the momentum is conserved when carts are collided on a slanting plane.

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a.) K 2=K 1 +GmM( r 21− r 11)=2.2×10 7J

b.) ​K 2 +GmM( r 11− r 21)=6.9×10 7 J

Applying Law of  Energy conservation :

K 1+U 1

=K 2+U 2

⇒K 1− r 1GmM

=K 2− r 2 GmM

where M=5.0×10 23kg,r1

=> R=3.0×10 6m and m=10kg

(a) If K 1​

=5.0×10 7J and r 2

=4.0×10 6 m, then the above equation leads to

K 2=K 1 +GmM (r 21− r 11)=2.2×10 7J

(b) In this case, we require K 2

=0 and r2

=8.0×10 6m, and solve for K 1:K 1

​=K 2 +GmM (r 11− r 21)=6.9×10 7 J

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Answer:

The velocity after 12s is 50.4m/s.

Explanation:

In acceleration formula make velocity the subject.

acceleration(a) = velocity(v)÷time(t)

V = 4.2m/s²×12s

V = 50.4m/s

Therefore the velocity after 12s is 50.4m/s.

The constant G in Newton's equation makes the units of measurement consistent was measured by Newton shows gravity to be a relatively huge force produces equilibrium makes the units of measurement consistent.

The three fundamental laws of classical mechanics known as Newton's laws of motion describe how an object's motion and the forces acting on it interact.

Newton's first law, an object will not change its motion unless a force acts on it.

Newton's second law, the force on an object is equal to its mass times its acceleration.

Newton's third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.

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Gravitational force is not related to other force because it is related to gravity.

Electrostatic force and magnetic force are link to each other because both have charges. Electrostatic force is type of force that is present between two electrically charged particles. They can either be a repulsive or attractive force.

So we can conclude that Gravitational force is not related to other force because it is related to gravity.

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The time needed for the hammer to reach the surface of the Earth is 3.54 s.

The time of motion is calculated as follows;

t = √(2h/g)

where;

t = √(2 x 10 / 1.6)

t = 3.54 s

Thus, the time needed for the hammer to reach the surface of the Earth is 3.54 s.

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Answer:thermal (heat) energy and radiant (light) energy.

Explanation:

Answer:

Answer: We use the chemical energy in fuels by burning them and transforming them into other types of energy: thermal energy, as when we burn fuel for heat; and kinetic energy, as when we burn gasoline to power our car's motion

Explanation:

hope it will help you

The launch velocity of the projectile is 13.28 m/s.

What is projectile motion?

The motion of an object thrown in the air under the force of gravity is known as projectile motion.

Since the object is launched horizontally, its initial velocity along the vertical direction is zero. From the second kinematic equation,

s=u*t+(1/2)at^2.

where s is the displacement, t is the time, u is the initial velocity and a is the acceleration. Since the height is decreasing, so it will be taken negative.

For the vertical motion, s=-19.3 m, a=-9.8 m/s^2 and u=0. Put the values in the above equation and solve it.

-19.3 = (0)*t+(1/2)*(-9.8)*t^2

19.3 = (1/2)*(9.8)*t^2

t=1.98 s

Since the velocity along the horizontal direction is constant, the displacement along the horizontal direction is given by the formula,

X=vt

where X is the horizontal displacement, v is the initial horizontal velocity and t is the time.

For the horizontal motion, X=26.3 m and t=1.98 s. Put the values in this equation and solve it.

26.3=v*(1.98)

v=13.28 m/s

The launch velocity is equal to the initial horizontal velocity, so it is equal to 13.28 m/s.

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Answer:

[tex] \tt \: mass \: = \frac{weight}{acceleration \: due \: to \: gravity} [/tex]

[tex] \longrightarrow \tt \: \frac{150}{10} [/tex]

[tex] \longrightarrow \boxed{ \tt{15 \: kg}}[/tex]

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3.375m/s is the final velocity of the car.

The final velocity depends on how large the acceleration is and the distance over which it acts.

Initial velocity of an object, you can multiply the acceleration due to a force by the time the force is applied and add it to the initial velocity to get the final velocity.

According to the question,

A toy car starts from the rest and accelerates

So the acceleration = 1.50m/s²

Time =  2.25s

[tex]x=x_{0} + vt[/tex]

[tex]x = 0 + ( 1.50m/s^2*2.25s)[/tex]

[tex]x = 3.375m/s[/tex]

The final velocity, of the car is 3.375 m/s.

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