Question 25
The drawings below show four objects that are acted upon by different forces The direction of the forces are represented by the arrows, and the units of the forces are in newtons (N) Based on Newton's law of
inertia, which object is moving at a steady speed, assuming that all four objects are moving on a frictionless surface?

Answer:

STOP CHETING

Answer:

C is the answer to your question

Explanation:

Have a great day!!! :)

Answer: the last one

Explanation: it pushes the lid off in a upward motion

Answer:

the last one

Explanation:

Can you mark me as BRAINLLIEST, PLEASE!!!!!!!!!!!!

(○’ω’○)          (ToT)  ↖(^ω^)↗  -_-z  ◑﹏◐  囧    ^O^  B-)

Answer:

A hygroscopic substance is one that readily attracts water from its surroundings, through either absorption or adsorption.

Explanation:

A hygroscopic substance is one that readily attracts water from its surroundings, through either absorption or adsorption.

Answer:

It’s a

Explanation:

Don’t actually put that i needed the points mb

Answer:

i don't know the correct answer

but this is what i found on web

Explanation:

Changes of state are physical changes in matter. They are reversible changes that do not change matter's chemical makeup or chemical properties. Processes involved in changes of state include melting, freezing, sublimation, deposition, condensation, and evaporation. Energy is always involved in changes of state.

Answer:

Explanation:

The speed of the horse can be found by using the formula

[tex]v = \frac{p}{m} \\ [/tex]

p is the momentum

m is the mass

From the question we have

[tex]v = \frac{1200}{313} \\ = 3.83386..[/tex]

We have the final answer as

Hope this helps you

i think the answer is h if am wrong am sorry

Explanation:

Answer:

The positions are  0.0194 m  and - 0.0194 m.

Explanation:

Given;

amplitude of the simple harmonic motion, A = 2.0 cm = 0.02 m

speed of simple harmonic motion is given as;

[tex]v = \omega \sqrt{A^2-x^2}[/tex]

the maximum speed of the simple harmonic motion is given as;

[tex]v_{max} = \omega A[/tex]

when the speed equal one fourth of its maximum speed

[tex]v =\frac{v_{max}}{4}[/tex]

[tex]\omega\sqrt{A^2-x^2} = \frac{\omega A}{4} \\\\\sqrt{A^2-x^2}= \frac{A}{4}\\\\A^2-x^2 = \frac{A^2}{16} \\\\x^2 = A^2 - \frac{A^2}{16} \\\\x^2 = \frac{16A^2 - A^2}{16} \\\\x^2 = \frac{15A^2}{16} \\\\x= \sqrt{\frac{15A^2}{16} } \\\\x = \sqrt{\frac{15(0.02)^2}{16} }\\\\x = 0.0194 \ m \ \ or\ - 0.0194 \ m[/tex]

Thus, the positions are  0.0194 m and - 0.0194 m.

The positions where the speed equals 1/4 of its maximum speed is mathematically given as

[tex]x \pm 0.0194[/tex]

Question Parameters:

an amplitude of 2.00 cm.

Generally, the equation for the speed of simple harmonic motion  is mathematically given as

[tex]v = w \sqrt{A^2-x^2}\\\\v=\frac{wA}{4}[/tex]

Therefore

[tex]w \sqrt{A^2-x^2}=\frac{wA}{4}\\\\x^2=A^2-A^2/16\\\\x=\sqrt{\frac{15A^2}{16}}[/tex]

[tex]x \pm 0.0194[/tex]

In conclusion, the positions are

[tex]x \pm 0.0194[/tex]

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Answer:

[tex]0.566\; \rm N[/tex] (assuming that while the block is moving, the friction on the block is constant.)

Explanation:

The mechanical energy of a system is the sum of its:

Friction does work on the block as the block moves up the ramp. The amount of energy that the block-string system has lost would be equal in size to the work that friction has done on the block. The size of the friction on the block could thus be computed.

Before this block was released, the block-spring system has no kinetic energy because there was no movement. Assume that the system has no gravitational potential energy at that moment, either. The only type of mechanical energy in this system at that moment would be elastic potential energy: [tex]8\; \rm J[/tex] according to the question.

The question states that the ramp is [tex]53.0^\circ[/tex] above horizontal. Therefore, after the block has traveled [tex]3.00\; \rm m[/tex] (along the ramp,) the height of this block would have increased by [tex]\Delta h = 3.00\; \rm \sin\left(53^\circ\right) \approx 2.39591\; \rm m[/tex]. Calculate the corresponding gain in gravitational potential energy:

[tex]\begin{aligned}& m \cdot g \cdot \Delta h\\ &\approx 0.200\; \rm kg& \\\ &\quad\times 9.81\; \rm N \cdot kg^{-1} \\ &\quad \times 2.39591\; \rm m \\ &\approx 4.70077\; \rm J\end{aligned}[/tex].

On the other hand, the question states that the speed of the block ([tex]m = 0.200\; \rm kg[/tex]) at that moment is [tex]v = 4.00\; \rm m \cdot s^{-1}[/tex]. Calculate the corresponding kinetic energy:

[tex]\begin{aligned}& \frac{1}{2}\, m \cdot v^{2} = 1.6\; \rm J \end{aligned}[/tex].

Because the block was already released, there should be no elastic potential energy in the spring.

Hence, the mechanical of the block-spring system would be approximately [tex]4.70077\; \rm J+ 1.6\; \rm J \approx 6.30077\; \rm J[/tex].

Approximate the amount of mechanical energy that is lost:

[tex]8\; \rm J - 6.30077\; \rm J \approx 1.69923\; \rm J[/tex].

In other words, when applied over [tex]3\; \rm m[/tex], the friction on this block would do approximately [tex]1.69923\; \rm J[/tex] of work. Approximate the size of that friction:

[tex]\begin{aligned}F &= \frac{W}{s} \\ &\approx \frac{1.69923\; \rm J}{3.00\; \rm m}\approx 0.566\; \rm N\end{aligned}[/tex].

Answer:B

Explanation: i took the test and i got it right

Answer:

Explanation:

Position of n the order fringe = n λ D / d

for n = 2

position = 2 λ D / d

λ = 710 nm , D = 1.1 m

d = .65 x 10⁻³

position 1 = 2 x 710 x 10⁻⁹ x 1.1 / .65 x 10⁻³

= 2403.07 x 10⁻⁶ m

= 2.403 x 10⁻³ m

= 2.403 mm .

For λ = 655 nm

position = 2 λ D / d

λ = 655 nm , D = 1.1 m

d = .65 x 10⁻³

position 2 = 2 x 655 x 10⁻⁹ x 1.1 / .65 x 10⁻³

= 2216.91 x 10⁻⁶ m

= 2.217 x 10⁻³ m

= 2.217 mm .

Difference between their position

= 2.403 - 2.217 = .186 mm .

The man is pulled upward by a net force of

∑ F = (86 kg) (1.35 m/s²) = 116.1 N

The net force is comprised of the tension T in the cable and the man's weight W, such that

∑ F = T - W = 116.1 N

T = (86 kg) (9.80 m/s²) + 116.1 N

T = 958.9 N ≈ 960 N

Answer:

a)t  = 1,43 s

b) V = 10,49 m/s

c) V₀ₓ = 10,49 m/s   ;    V₀y = 14,01 m/s

d) Vf = 17,5 m/s

Explanation:

According to the problem statement

V₀ = V₀ₓ    and  V₀y = 0

And at the end of the movement t = ?  the distance y = 10 m

Therefore as

h = V₀y - (1/2)*g*t²

Vertical distance y = h = 10 = V₀y*t - 0,5 (-9,8)*t²

10 = 4,9*t²

t² = 10/4,9    ⇒  t² = 2,04 s

t  = 1,43 s

a) 1,43 s is the time of movement

b) V₀ = V₀ₓ        V₀y = 0     and  V₀ₓ = Vₓ     ( constant )

Just before touching the ground, the horizontal distance is

hd = 15 = Vₓ * t

Then  15 /1,43 = Vₓ = V₀ₓ

Vₓ = 10,49 m/s

Then initial speed is V = 10,49 m/s    since V₀y = 0

Vf² = Vₓ² + Vy²

Vyf = V₀y - g*t

Vyf =  0 - 9,8 *1,43

Vyf = - 14,01 m/s

And finally the speed when the projectile strike the ground is:

Vf² = Vₓ² + Vy²

Vf = √ (10,49)² + (14,01)²

Vf = 17,50 m/s

Answer:

20 m/s^2

Explanation:

Use the formula Vf=Vi+at

Vf=100

Vi=0

a=?

t=5

Then solve for (a)

Electrical power can be defined as the product of Voltage and current. Thus, P = IV. Hence, Option D is the correct answer.

Thus, the formula for electric power,

P = IV watt

Where,

P - Power (Watt)

I - Current (Ampere)

V - Voltage (volts)

Electrical power can be defined as the product of Voltage and current. Thus, P = IV. Hence, Option D is the correct answer.

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Answer:

The final velocity is 20 m/s.

Explanation:

Constant Acceleration Motion

It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.

Being a the constant acceleration, vo the initial speed, and t the time, the final speed can be calculated as follows:

[tex]v_f=v_o+at[/tex]

The provided data is: vo=10 m/s, [tex]a=5\ m/s^2[/tex], t=2 s. The final velocity is:

[tex]v_f=10~m/s+5\ m/s^2\cdot 2\ s[/tex]

[tex]v_f=20\ m/s[/tex]

The final velocity is 20 m/s.

Explanation:It is given that,

Mass of bumper car, m₁ = 202 kg

Initial speed of the bumper car, u₁ = 8.5 m/s

Mass of the other car, m₂ = 355 kg

Initial velocity of the other car is 0 as it at rest, u₂ = 0

Final velocity of the other car after collision, v₂ = 5.8 m/s

Let p₁ is momentum of of 202 kg car, p₁ = m₁v₁

Using the conservation of linear momentum as :

p₁ = m₁v₁ = -342 kg-m/s

So, the momentum of the 202 kg car afterwards is 342 kg-m/s. Hence, this is the required solution.

Answer:

1717 kg*m/s

Explanation:

found it put it in on acellus and it was right

Answer:

F = 288 [N]

Explanation:

To solve this problem we must use the following equation of kinematics and find the value of acceleration.

[tex]v_{f}^{2} =v_{o}^{2} +2*a*x[/tex]

where:

Vf =  final velocity = 6 [m/s]

Vo =  initial velocity = 0 (starting from rest)

a = acceleration [m/s²]

x = distance = 5 [m]

Now replacing, we have:

[tex](6)^{2}=0+(2*a*5)\\36=10*a\\a = 3.6 [m/s^{2}][/tex]

Since we already have the value of acceleration, we can use Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

ΣF = m*a

[tex]F =80*3.6\\F = 288 [N][/tex]

Answer:

656.5 m/s

Explanation:

V = ? U = 0 m/s a = 5.05 m/s2 S = 65.0 m

V = u + 2as

V = 0 + 2× 5.05 × 65

V = 656.5 m/s

Answer:

Momentum

Explanation:

In order for an object's momentum to change,a force must be applied for a period of time

Answer:25 N

Explanation:

in the case shown below, the 1 kg rock rides on a horizontal disk that rotates at constant speed 5m/s is 25N

Speed is the ratio of distance with respect to the time in which the distance was covered. Speed is a scalar quantity as it does not have magnitude only have direction

The formula of speed can be represented as s=d/t, Where, s is the speed in m.s-1, d is the distance traveled in m, t is the time taken in s

Uniform speed is defined when the object covers equal distance at equal time intervals, variable speed is defined as when the object covers a different distance at equal intervals of times.

Average speed is defined as the total distance travelled by an object to the total time taken by the object.

Instantaneous speed is defined as when the object is move with variable speed, then the speed at any instant of time is known as instantaneous speed.

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Answer:

s = 3.26 m

Explanation:

Given that,

Water leaves at a speed of 8 m/s

We need to find the maximum height that steelhead can jump. Let it can jump to a height of h.

At maximum height, final speed is equal to 0. We can use third equation of motion to find the maximum height.

[tex]v^2-u^2=2as[/tex]

a = -g

[tex]-u^2=-2gs\\\\s=\dfrac{u^2}{2g}\\\\s=\dfrac{(8)^2}{2\times 9.8}\\\\=3.26\ m[/tex]

Hence, the maximum height is 3.26 m.

Answer:

Approximately 1.02 seconds

Explanation:

Use the final velocity (vf) formula for a uniformly accelerated movement under "g" (acceleration of gravity):

[tex]v_f=v_i+g\,*\,t[/tex]

in our case:

[tex]10=0+9.8\,*\,t\\t=10/9.8\\t\approx 1.02\,\,sec[/tex]

Answer:

jack has done more work pulling more weight and Jill has more power.

Explanation: