A ball is falling toward the ground. Which is true?

A block slides to a stop as it goes 47 m across a level floor in a time of 6.35 s. a) What was the initial velocity? b) What is

UNO [17]

Answer :

(a) The initial velocity is, 14.8 m/s

(b) The acceleration is, $-2.33m/s^2$

Explanation :

By the 1st equation of motion,

$v=u+at$ ...........(1)

where,

v = final velocity = 0 s

u = initial velocity

t = time = 6.35 s

a = acceleration

The equation 1 will be:

$0=u+a(6.35}$

$u=-6.35a$ ..........(2)

By the 2nd equation of motion,

$s=ut+\frac{1}{2}at^2$ ...........(3)

where,

s = distance = 47 m

Now substitute equation 2 in 3, we get:

$47=(-6.35a)\times (6.35)+\frac{1}{2}\times a\times (6.35)^2$

By solving the term, we get:

$a=-2.33m/s^2$

The acceleration is, $-2.33m/s^2$

Now we have to calculate the initial velocity.

Using equation 2, we gte:

$u=-6.35a$

$u=-6.35s\times (-2.33m/s^2)$

$u=14.8m/s$

The initial velocity is, 14.8 m/s

5 0

3 years ago

A car is traveling on a straight, level road under wintry conditions. Seeing a patch of ice ahead of her, the driver of the car

dangina [55]

Her magnitude of deceleration on the ice would be 15.126m/s

5 0

3 years ago

How does the study of social sciences affect our society?

Kruka [31]

Explanation:

study of social sciencake us an effcient citizen of a democracy ,and it also help us to solve the practical problem in our daily life.

8 0

4 years ago

You are working with a team that is designing a new roller coaster-type amusement park ride for a major theme park. You are pres

Dominik [7]

Answer:

<em>The required constant friction force for the last 20 m is 6,862.8 N</em>

Explanation:

<u>Energy Conversion</u>

There are several ways the energy is manifested in our physical reality. Some examples are Kinetic, Elastic, Chemical, Electric, Potential, Thermal, Mechanical, just to mention some.

The energy can be converted from one form to another by changing the conditions the objects behave. The question at hand states some types of energy that properly managed, will make the situation keep under control.

Originally, the m=220 kg car is at (near) rest at the top of a h=101 m tall track. We can assume the only energy present at that moment is the potential gravitational energy:

$E_1=mgh=220\cdot 9.8\cdot 101=217,756\ J$

For the next x1=230 m, a constant friction force Fr1=350 N is applied until it reaches ground level. This means all the potential gravitational energy was converted to speed (kinetic energy K1) and friction (thermal energy W1). Thus

$E_1=K_1+W_1$

We can compute the thermal energy lost during this part of the motion by using the constant friction force and the distance traveled:

$W_1=F_{r1}\cdot x_1=350\cdot 230=80,500\ J$

This means that the kinetic energy that remains when the car reaches ground level is

$K_1=E_1-W_1=217,756\ J-80,500\ J=137,256\ J$

We could calculate the speed at that point but it's not required or necessary. That kinetic energy is what keeps the car moving to its last section of x2=20 m where a final friction force Fr2 will be applied to completely stop it. This means all the kinetic energy will be converted to thermal energy:

$W_2=F_{r2}\cdot x_2=137,256$

Solving for Fr2

$\displaystyle F_{r2}=\frac{137,256}{20}=6,862.8\ N$

The required constant friction force for the last 20 m is 6,862.8 N

3 0

3 years ago

A pulley with the radius of 10.0 cm was connected to a motor with a massless

kogti [31]

Answer:

(i) -556 rad/s²

(ii) 17900 revolutions

(iii) 11250 meters

(iv) -55.6 m/s²

(v) 18 seconds

Explanation:

(i) Angular acceleration is change in angular velocity over time.

α = (ω − ω₀) / t

α = (10000 − 15000) / 9

α ≈ -556 rad/s²

(ii) Constant acceleration equation:

θ = θ₀ + ω₀ t + ½ αt²

θ = 0 + (15000) (9) + ½ (-556) (9)²

θ = 112500 radians

θ ≈ 17900 revolutions

(iii) Linear displacement equals radius times angular displacement:

s = rθ

s = (0.100 m) (112500 radians)

s = 11250 meters

(iv) Linear acceleration equals radius times angular acceleration:

a = rα

a = (0.100 m) (-556 rad/s²)

a = -55.6 m/s²

(v) Angular acceleration is change in angular velocity over time.

α = (ω − ω₀) / t

-556 = (0 − 15000) / t

t = 27

t − 9 = 18 seconds

8 0

3 years ago