Ask question

Ask question

All categories

2 years ago

11

An object moving at a velocity of 32m/s slows to a stop in 4 seconds. What was its acceleration?

2 answers:

Romashka [77]2 years ago

6 0

Answer:

8m/s

Explanation:

a=d/t

a=32/4

a=8 m/s

LiRa [457]2 years ago

5 0

Answer:

-8m/s^2

Explanation:

initial velocity(u)=32m/s

Final velocity(v)=0

Time(t)=4seconds

Acceleration(a)=?

v=u+axt

0=32+ax4

-32=4a

a=-32/4

a=-8m/s^2

You might be interested in

A hot piece of iron is thrown into the ocean and its temperature eventually stabilizes. Which of the following statements concer

777dan777 [17]

Answer:

E. The ocean gains more entropy than the iron loses.

Explanation:

When there is a spontaneous process , entropy of the system increases . Here hot iron is losing entropy and ocean is gaining entropy . Net effect will be gain of entropy . That means entropy gained by ocean is more than entropy lost by iron .

Hence option E is correct .

8 0

2 years ago

Briana swings a ball on the end of a rope in a circle. The rope is 1.5 m long. The ball completes a full circle every 2.2 s. Wha

schepotkina [342]

The radius of the circular path is 1.5 m.

The circumference is then
$1.5\ m*2\pi=3\pi\ m$

The ball moves 3π m every 2.2 s, so the speed is
$\frac{3\pi\ m}{2.2\ s}\approx 4.3\ m/s$

9 0

2 years ago

Read 2 more answers

A cannon is mounted on a tower above a wide, level field. The barrel of the cannon is 20 m above the ground below. A cannonball

OLga [1]

<u>Answer:</u>

Cannonball will be in flight before it hits the ground for 2.02 seconds

<u>Explanation:</u>

Initial height from ground = 20 meter.

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this the velocity of body in vertical direction = 0 m/s, acceleration = 9.8 $m/s^2$ , we need to calculate time when s = 20 meter.

Substituting

$20=0*t+\frac{1}{2} *9.8*t^2\\ \\ t = 2.02 seconds$

So it will take 2.02 seconds to reach ground.

5 0

2 years ago

A physics student shoves a 0.50-kg block from the bottom of a frictionless 30.0° inclined plane. The student performs 4.0 j of w

Musya8 [376]

Answer:1.63 m

Explanation:

Given

mass of block $m=0.5 kg$

inclination $\theta =30^{\circ}$

Amount of work done $W=4 J$

block slides a distance s along the Plane

Work done =change in Potential Energy

Increase in height of block is $s\sin \theta$

Change in Potential Energy $=mg(\sin \theta -0)$

$\Delta P.E.=0.5\times 9.8\times s\sin 30$

$4=0.5\times 9.8\times s\sin 30$

$s=\frac{4}{2.45}$

$s=1.63 m$

5 0

2 years ago

Suppose a rectangular piece of aluminum has a length D, and its square cross section has the dimensions W XW, where D (W x W) to

Ludmilka [50]

Answer:

R₂ / R₁ = D / L

Explanation:

The resistance of a metal is

R = ρ L / A

Where ρ is the resistivity of aluminum, L is the length of the resistance and A its cross section

We apply this formal to both configurations

Small face measurements (W W)

The length is

L = W

Area

A = W W = W²

R₁ = ρ W / W² = ρ / W

Large face measurements (D L)

Length L = D= 2W

Area A = W L

R₂ = ρ D / WL = ρ 2W / W L = 2 ρ/L

The relationship is

R₂ / R₁ = 2W²/L

6 0

2 years ago