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Sedaia [141]
2 years ago
10

A shopping cart given an initial velocity of 2.0 m/s undergoes a constant acceleration of 3.0 m/s2. What is the magnitude of the

cart’s displacement after the first 4.0 s of its motion?
Physics
1 answer:
Tema [17]2 years ago
6 0

Answer:

32 m

Explanation:

initial velocity (u) = 2 m/s

accceleration (a) = 3 m/s^{2}

time (t) = 4 s

What is the magnitude of the cart’s displacement (s) after the first 4.0 s of its motion?

from the equation of motion, displacement (s) = ut + 0.5at^{2}

where

  • u is the initial velocity
  • t is the time
  • a is the acceleration

inserting all required values into  (s) = ut + 0.5at^{2}          

(s) = 2x4 + 0.5x3x4^{2}

s = 8 + 24

s = 32 m

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You may have noticed runaway truck lanes while driving in the mountains. These gravel-filled lanes are designed to stop trucks t
Sladkaya [172]

Answer:

0.767

Explanation:

The work done on the truck by the frictional drag force is given by

W=-Fd

where

F is the magnitude of the frictional force

d = 38.0 m is the maximum displacement allowed for the truck

The negative sign is due to the fact that the force of friction is opposite to the motion of the truck

The force of friction can also be written as:

F=\mu mg

where

\mu is the coefficient of kinetic friction between the truck and the lane

m is the mass of the truck

g is the acceleration of gravity

So we can rewrite the work done as

W=-\mu mg d (1)

According to the work-energy theorem, the work done by friction is equal to the change in kinetic energy of the truck:

W=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2 (2)

where

v = 0 is the final velocity of the truck

u = 23.9 m/s is the initial velocity of the truck

By combining (1) and (2) we get

-\frac{1}{2}mu^2 = -\mu mg d

And solving for \mu, we find the minimum coefficient of kinetic friction able to stop the truck in a distance d:

\mu = \frac{u^2}{2gd}=\frac{23.9^2}{2(9.8)(38.0)}=0.767

7 0
2 years ago
A granite monument has a volume of 25,365.4 cm3. The density of granite is 2.7 g/cm3. Use this information to calculate the mass
LiRa [457]
As density = mass/volume

So

Mass = density *volume

Mass = 25,365.4 * 2.7 = 68,486.58 g

<span>Mass of the granite monument to the nearest tenth = 68,485.6 g</span>

5 0
2 years ago
A highly charged piece of metal (with uniform potential throughout) tends to spark at places where the radius of curvature is sm
k0ka [10]

Answer:

look it up

Explanation:

8 0
2 years ago
A 1.50-m cylinder of radius 1.10 cm is made of a complicated mixture of materials. Its resistivity depends on the distance x fro
MArishka [77]

Answer:

Resistance = 3.35*10^{-4} Ω

Explanation:

Since resistance R = ρ\frac{L}{A}

whereas \rho(x) = a + bx^2

resistivity is given for two ends. At the left end resistivity is 2.25* 10^{-8} whereas x at the left end will be 0 as distance is zero. Thus

2.25*10^{-8} = a + b(0)^2\\ 2.25*10^{-8} = a + 0 \\2.25*10^{-8} = a

At the right end x will be equal to the length of the rod, so x = 1.50\\8.50*10^{-8} = (2.25*10^{-8}) + ( b* (1.50)^2 )\\8.50*10^{-8} - (2.25*10^{-8}) = b*2.25\\\frac{6.25*10^{-8}}{2.25}  = b\\b = 2.77 *10^{-8}

Thus resistance will be R = ρ\frac{L}{A}

where A = π r^2

so,

R = \frac{8.50*10^{-8} * 1.50}{3.14*(1.10*10^{-2})^2} \\R=3.35 * 10 ^{-4}

6 0
2 years ago
To determine the height of a flagpole, Abby throws a ball straight up and times it. She sees that the ball goes by the top of th
Kryger [21]

Answer:

H = 10.05 m

Explanation:

If the stone will reach the top position of flag pole at t = 0.5 s and t = 4.1 s

so here the total time of the motion above the top point of pole is given as

\Delta t = 4.1 - 0.5 = 3.6 s

now we have

\Delta t = \frac{2v}{g}

3.6 = \frac{2v}{9.8}

v = 17.64 m/s

so this is the speed at the top of flag pole

now we have

v_f - v_i = at

17.64 - v_i = (-9.8)(0.5)

v_i = 22.5 m/s

now the height of flag pole is given as

H = \frac{v_f + v_i}{2}t

H = \frac{22.5 + 17.64}{2} (0.5)

H = 10.05 m

5 0
2 years ago
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