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Sergio [31]
2 years ago
13

If an object is accelerating, it is traveling the same distance for each time interval of its motion

Physics
2 answers:
Elenna [48]2 years ago
6 0
The answer is false :)
Anika [276]2 years ago
4 0

Answer:

false

Explanation:

if an object is accelerating the object will not travel the same distance every time interval

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Andrews [41]

whats the restraunt called

6 0
2 years ago
A worker lifts a 20.0-kg bucket of concrete from the ground up to the top of a 25.0-m tall building. The bucket is initially at
yuradex [85]

Answer:

Minimum work = 5060 J

Explanation:

Given:

Mass of the bucket (m) = 20.0 kg

Initial speed of the bucket (u) = 0 m/s

Final speed of the bucket (v) = 4.0 m/s

Displacement of the bucket (h) = 25.0 m

Let 'W' be the work done by the worker in lifting the bucket.

So, we know from work-energy theorem that, work done by a force is equal to the change in the mechanical energy of the system.

Change in mechanical energy is equal to the sum of change in potential energy and kinetic energy. Therefore,

\Delta E=\Delta U+\Delta K\\\\\Delta E= mgh+\frac{1}{2}m(v^2-u^2)

Therefore, the work done by the worker in lifting the bucket is given as:

W=\Delta E\\\\W=mgh+\frac{1}{2}m(v^2-u^2)

Now, plug in the values given and solve for 'W'. This gives,

W=(20\ kg)(9.8\ m/s^2)(25\ m)+\frac{1}{2}(20\ kg)(4^2-0^2)\ m^2/s^2\\\\W=4900\ J +160\ J\\\\W=5060\ J

Therefore, the minimum work that the worker did in lifting the bucket is 5060 J.

7 0
3 years ago
How does inertia affect someone who isn't wearing a seatbelt
Paul [167]
Basically, when someone is resting in an accelerated vehicle without restraint from a seatbelt, the force of stopping the vehicle will be when inertia occurs, and that force of the vehicle coming to a stop will affect the passenger (without a seatbelt/restraint from another force or object) greatly by throwing them.
For example;
If I were to be riding in a vehicle (without a seatbelt) that's accelerating at 40 m/s^2 and it suddenly gets slammed on the breaks, I will be thrown forward from inside the vehicle.

I hope this helps!
7 0
3 years ago
A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.76 m/s2 for t1 = 20 s. At that point the
alex41 [277]

Answer:

(a)v_1 = a_1t_1 = 1.76 t_1

(b) It won't hit

(c) 110 m

Explanation:

(a) the car velocity is the initial velocity (at rest so 0) plus product of acceleration and time t1

v_1 = v_0 + a_1t_1 = 0 +1.76t_1 = 1.76t_1

(b) The velocity of the car before the driver begins braking is

v_1 = 1.76*20 = 35.2m/s

The driver brakes hard and come to rest for t2 = 5s. This means the deceleration of the driver during braking process is

a_2 = \frac{\Delta v_2}{\Delta t_2} = \frac{v_2 - v_1}{t_2} = \frac{0 - 35.2}{5} = -7.04 m/s^2

We can use the following equation of motion to calculate how far the car has travel since braking to stop

s_2 = v_1t_2 + a_2t_2^2/2

s_2 = 35.2*5 - 7.04*5^2/2 = 88 m

Also the distance from start to where the driver starts braking is

s_1 = a_1t_1^2/2 = 1.76*20^2/2 = 352

So the total distance from rest to stop is 352 + 88 = 440 m < 550 m so the car won't hit the limb

(c) The distance from the limb to where the car stops is 550 - 440 = 110 m

8 0
3 years ago
Please help me! It states constant speed but moving in circular motion. I know there is acceleration due to a rate of change of
Morgarella [4.7K]
Direction of resultant force in circular motion is directed towards the center of the circle. Hence vector B is the direction of the resultant force.
8 0
3 years ago
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