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Anastasy [175]
3 years ago
15

For a certain interval of time, an object is acted upon by a constant non-zero force. Which of the following statements is true

for this interval of time?
Physics
1 answer:
Serjik [45]3 years ago
5 0

Missing options:

a. The object is accelerating.

b. The object's velocity can only increase.

c. The object is at rest.

d. The object is moving with constant velocity.

e. The object's velocity changes.

Answer:

a. The object is accelerating.

e. The object's velocity changes.

Explanation:

Let's answer by using Newton's second law:

F = ma

where

F is the net force on the object

m is the mass of the object

a is the acceleration

From the equation, we see that if the force on an object is non-zero, then the object is for sure accelerating, since a is different from zero.

So, option A is correct.

Now, we also recall the definition of acceleration, which is the rate of change of velocity:

a=\frac{\Delta v}{\Delta t}

where \Delta v is the change in velocity in the time interval \Delta t. This can be rewritten as

\Delta v = a \Delta t

Here, since the acceleration is non-zero, it follows that also \Delta v\neq 0: which means that the velocity of the object is changing.

So, option E is also correct.

The other options are wrong because:

b. The object's velocity can only increase.  --> false, in fact acceleration can also be negative, and in that case the velocity would be decreasing

c. The object is at rest.  --> false, since the velocity of the object is changing

d. The object is moving with constant velocity.  --> false, since the velocity of the object is changing

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A cheetah spotted a Gazelle. The cheetah runs at its top speed of 30 m/s for 15 seconds. Durning this time , a gazelle ,160 m fr
Xelga [282]

Answer:

A) 450 m

B) 27 m/s

C) 81 m, 243 m

D) Gazelle

Explanation:

A)

Since, the Cheetah is running at constant speed. Therefore, we use the equation:

s₁ = v₁t

where,

s₁ = distance covered by Cheetah = ?

v₁ = speed of Cheetah = 30 m/s

t = time taken = 15 s

Therefore,

s₁ = (30 m/s)(15 s)

<u>s₁ = 450 m</u>

<u></u>

B)

For final speed of Gazelle at the end of 6 s acceleration time we use 1st equation of motion:

Vf = Vi + at

where,

Vf = Final Speed of Gazelle at the end of 6 s = ?

Vi = Initial Speed of Gazelle = 0 m/s (Since, Gazelle is initially at rest)

t = time taken = 6 s

a = acceleration = 4.5 m/s²

Therefore,

Vf = (0 m/s) + (4.5 m/s²)(6 s)

<u>Vf = 27 m/s</u>

C)

For the distance covered by Gazelle at the end of 6 s acceleration time we use 2nd equation of motion:

s₂ = Vi t + (0.5)at²

where,

Vi = Initial Speed of Gazelle = 0 m/s (Since, Gazelle is initially at rest)

t = time taken = 6 s

a = acceleration = 4.5 m/s²

Therefore,

s₂ = (0 m/s)(6 s) + (0.5)(4.5 m/s²)(6 s)²

<u>s₂ = 81 m</u>

<u></u>

Now, for the distance covered during the last 9 s at constant velocity Vf, we use equation:

s₃ = Vf t

where,

s₃ = distance covered by Gazelle in last 9 s = ?

t = time = 9 s

Therefore:

s₃ = (27 m/s)(9 s)

<u>s₃ = 243 m</u>

D)

We know that, at the end of 15 seconds:

Distance covered by Cheetah = s₁ = 450 m

Distance Covered by Gazelle = s₂ + s₃ = 81 m + 243 m = 324 m

If we take the initial position of Cheetah as origin. Then the positions of both Gazelle and Cheetah with respect to origin will be:

Position of Cheetah = 450 m ahead of origin

Position of Gazelle = 324 m + 160 m = 484 m (Since, Gazelle was initially 160 m ahead of Cheetah)

<u>Hence, it is clear that Gazelle is ahead at the end of 15 s.</u>

5 0
3 years ago
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Ne4ueva [31]

Answer:

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A 1.150 kg air-track glider is attached to each end of the track by two coil springs. It takes a horizontal force of 0.900 N to
shutvik [7]

k = 5.29

a = 0.78m/s²

KE = 0.0765J

<u>Explanation:</u>

Given-

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distance, x = 0.17m

(a) Effective spring constant, k = ?

Force = kx

0.9 = k X0.17

k = 5.29

(b) Maximum acceleration, m = ?

We know,

Force = ma

0.9N = 1.15 X a

a = 0.78 m/s²

c) kinetic energy, KE of the glider at x = 0.00 m.

The work done as the glider was moved = Average force * distance

This work is converted into kinetic energy when the block is released. The maximum kinetic energy occurs when the glider has moved 0.17m back to position x = 0  

As the glider is moved 0.17m, the average force = ½ * (0 + 0.9)

Work = Kinetic energy

KE = 0.450 * 0.17

KE = 0.0765J

4 0
3 years ago
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