6. The volume of a water tank is 5 m × 4 mx 2 m. If the tank is half-filled with water, calculate the pressure exerted at the bottom of the tank.
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If the current that is circuit is tripled,  the power is increased by a factor of 9. Option D

The term power refers to the rate of doing work. Now if we have to obtain the work done by the use of the formula; P = I^2R

Now if the current that is circuit is tripled, the power is now obtained by;

P = (3I)^2R = 9I^2R hence the power is increased by a factor of 9.

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Answer:

The efficiency of the light bulb is 30%.

Explanation:

Efficiency = output/input × 100%

Data

Energy input = 100J

Energy output = 30J

Efficiency = ?

Efficiency = Energy output/Energy input × 100%

Efficiency = 30J/100J × 100%

Efficiency = 0.3 × 100%

Efficiency = 30%

Therefore the efficiency of the light bulb is 30%.

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.

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:

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

Explanation:

Hope this helps

Ultraviolet rays are Electromagnetic waves that have higher frequencies than visible light, so they carry more energy.

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Answer:

F = K Q1 Q2 / R^2       force between 2 charged partices

F2 / F1 = (R1 / R2)^2 = (1 / 2.69)^2 = .139

F2 = .139 F1

The time in which the animal B overtake animal A according to the motion map is at 2 seconds.

This is referred to as one-dimensional plots built off of a position line and depicts velocity and acceleration.

The time in which Animal B is ahead of animal A is at 2 seconds which therefore makes it the most appropriate choice.

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Answer:

the gravitational force

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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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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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.

Given the electronegativity of O is 3.5 and H is 2.1,  the bond between the two atoms will be classified as a polar covalent bond.

According to the difference in electronegativity (ΔEN).

Oxygen has an electronegativity of 3.5 and hydrogen of 2.1. The difference in electronegativity is:

ΔEN = EN(O) - EN(H) = 3.5 - 2.1 = 1.4

Given the electronegativity of O is 3.5 and H is 2.1,  the bond between the two atoms will be classified as a polar covalent bond.

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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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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].

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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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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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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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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【Answer】Therefore, the mass in said substance is 375 kilograms.

                               [tex]\red{ {\hspace{50 pt}\above 1.2pt}\boldsymbol{\mathsf{Procedure}}{\hspace{50pt}\above 1.2pt}}[/tex]

This is an exercise on fluids and their fundamental characteristics.

We start to solve, obtaining the data:

                                   Conversion from liters to m³

                                      [tex]\boldsymbol{1500\not{l}*\dfrac{1 \ m^{3} }{1000\not{l} }=1.5 \ m^{3} }[/tex]

To calculate mass: multiply density by volume.

                                       [tex]\boldsymbol{m=d*v \ \ \to \ \ \ Formula}[/tex]

We clear our data in the formula:

                                       [tex]\boldsymbol{m=250\dfrac{kg}{\not{m^{3}}}*1.5\not{m^{3}} }[/tex]

                                        [tex]\boldsymbol{m=375 \ kg}[/tex]

Answer:

9.8 m/s

Explanation:

If these objects are in free fall, meaning that their downward movement is only due to the force of gravity, then acceleration due to gravity is about 9.8 metres per second per second.

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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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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The horizontal force that the seat must be pulled so that the ropes form an angle of 35o with respect to the vertical is 345.2 N.

The horizontal force that the seat must be pulled so that the ropes form an angle of 35o with respect to the vertical is calculated as;

F = Wcosθ

where;

F = (43 x 9.8) x cos(35)

F = 345.2 N

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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

Answer:

The cliff is 74.4m.

Explanation:

The reading above clearly state that a ball is dropped straight down with the speed of 9.3m/s and hit the ground exactly at a time of 8.0s after you released it. And it is asking for the distance of the cliff.

In speed formula make distance the subject.

Therefore Distance = speed × time

Distance = 9.3mls × 8.0s = 74.4m

Therefore the cliff is 74.4m.

The circuit change when the wire is added will see a short circuit occur and makes bulbs 1 and 2 turn off but keeps bulbs 3 and 4 lit. Option D. This is further explained below.

Generally, Electronic circuits consist of a series of interconnected parts that form a closed loop through which electricity may flow.

In conclusion, If two wires are linked together, a short circuit will develop, cutting power to bulbs 1 and 2. But there is no impact on bulbs 3 and 4. There is no problem with bulbs 3 and 4.

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