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yulyashka [42]
3 years ago
11

Using this formula Vj = V; + at, If a vehicle starts from rest

Physics
1 answer:
Furkat [3]3 years ago
5 0

Answer:

<em>The final speed of the vehicle is 36 m/s</em>

Explanation:

<u>Uniform Acceleration</u>

When an object changes its velocity at the same rate, the acceleration is constant.

The relation between the initial and final speeds is:

v_f=v_o+a.t

Where:

vf  = Final speed

vo = Initial speed

a   = Constant acceleration

t   = Elapsed time

The vehicle starts from rest (vo=0) and accelerates at a=4.5 m/s2 for t=8 seconds. The final speed is:

v_f=0+4.5*8

v_f=36\ m/s

The final speed of the vehicle is 36 m/s

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The ability of sculptural material to resist forces of pressure, like gravity, is called its __________.
Vitek1552 [10]
I think the answer would be tensile, I’m sorry if it’s wrong
3 0
3 years ago
A stone is dropped from the upper observation deck of a tower, 750 m above the ground. (assume g = 9.8 m/s2.) (a) find the dista
Gekata [30.6K]
(a) The stone moves by uniform accelerated motion, with constant acceleration g=9.81 m/s^2 directed downwards, and its initial vertical position at time t=0 is 750 m. So, the vertical position (in meters) at any time t can be written as
y(t)= y_0 -  \frac{1}{2}gt^2= 750 - 4.9 t^2

(b) The time the stone takes to reach the ground is the time at which the vertical position of the stone becomes zero: y(t)=0. So, we can write
750-4.9 t^2 = 0
from which we find the time t after which the stone reaches the ground:
t= \sqrt{\frac{750 m}{4.9 m/s^2 }}= 12.37 s

(c) The velocity of the stone at time t can be written as
v(t) = -gt
because it is an accelerated motion with initial speed zero. Substituting t=12.37 s, we find the final velocity of the stone:
v(12.37 s)=-(9.81 m/s^2)(12.37 s)=-121.3 m/s

(d) if the stone has an initial velocity of v_0 = 6 m/s, then its law of motion would be
y(t)=y_0 - v_0t -  \frac{1}{2}gt^2
and we can find the time it needs to reach the ground by requiring again y(t)=0:
0=750 - 6t - 4.9 t^2
which has two solutions: one is negative so we neglect it, while the second one is t=11.78 s, so this is the time after which the stone reaches the ground.

5 0
3 years ago
Which statements describe how a machine can help make work easier? Select two options.
ad-work [718]

Answer:

  • It can put out more force than the input force by decreasing the distance through which force is applied.
  • It can apply a force to an object in a different direction than the force applied to the machine.

Explanation:

Assuming the machine is a simple machine, such as a lever or pulley, the machine will not do any more work than is put into it. However, it can change the magnitude or direction (or both) of the input force required to achieve a particular output force.

Because the machine does not increase the work done, if the output force is greater, the output distance must be less.

The applicable observations are ...

  • It can put out more force than the input force by decreasing the distance through which force is applied.
  • It can apply a force to an object in a different direction than the force applied to the machine.

6 0
3 years ago
A 28.0 kg child plays on a swing having support ropes that are 2.30 m long. A friend pulls her back until the ropes are 45.0 ∘ f
Dimas [21]

Answer

A)184.9J

B)=3.63m/s

C) Zero

Explanation:

A)potential energy of the child at the initial position, measured relative the her potential energy at the bottom of the motion, is

U=Mgh

Where m=28kg

g= 9.8m/s

h= difference in height between the initial position and the bottom position

We are told that the rope is L = 2.30 m long and inclined at 45.0° from the vertical

h=L-Lcos(x)= L(1-cosx)=2.30(1-cos45)

=0.674m

Her Potential Energy will now

= 28× 9.8×0.674

=184.9J

B)we can see that at the bottom of the motion, all the initial potential energy of the child has been converted into kinetic energy:

E= 0.5mv^2

where

m = 28.0 kg is the mass of the child

v is the speed of the child at the bottom position

Solving the equation for v, we find

V=√2k/m

V=√(2×184.9/28

=3.63m/s

C)we can find work done by the tension in the rope is given using expresion below

W= Tdcosx

where W= work done

T is the tension

d = displacement of the child

x= angle between the directions of T and d

In this situation, we have that the tension in the rope, T, is always perpendicular to the displacement of the child, d. x= 90∘ and cos90∘=0 hence, the work done is zero.

6 0
3 years ago
What is the potential energy of a 2kg object placed 6m above<br> the surface of the Earth?
zalisa [80]

Answer:117.6joules

Explanation:

Mass(m)=2kg

Height(h)=6m

Acceleration due to gravity(g)=9.8m/s^2

Potential energy(PE)=?

PE=m x g x h

PE=2 x 9.8 x 6

PE=117.6joules

8 0
3 years ago
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