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Sidana [21]
3 years ago
13

You are in a planet where the acceleration due to gravity is known to be 3.28 m/s^2. You drop a ball and record that the ball ta

kes 2.40 s to reach the ground. How fast is the ball moving just as it lands? Neglect air resistance.
A.) 1.12 m/s

B.) 7.87 m/s

C.) 3.55 m/s

D.) 2.81 m/s
Physics
1 answer:
notsponge [240]3 years ago
5 0
B) 7.87 m/s

The gravitational pull is the rate of change of velocity which is the acceleration. Formula for acceleration is;
a =  \frac{final \: velocity - initial \: velocity}{time \: taken}

Given:

• Initial velocity = 0m/s; I dropped the ball, and didn't throw it, so it was at rest firstly
• Time taken = 2.40s
• Acceleration = 3.28m/s^2

We're require to find the final velocity, at which the ball hit the ground with. Ignoring air resistance, keep in mind that the velocity of an object increases as it comes closer to the ground.

3.28  =  \frac{final \: velocity - 0}{2.40}
3.28 \times 2.40 = final \: velocity
final \: velocity = 7.872 \frac{m}{ {s} }

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pantera1 [17]

(8) A car starting with a speed <em>v</em> skids to a stop over a distance <em>d</em>, which means the brakes apply an acceleration <em>a</em> such that

0² - <em>v</em>² = 2 <em>a</em> <em>d</em> → <em>a</em> = - <em>v</em>² / (2<em>d</em>)

Then the car comes to rest over a distance of

<em>d</em> = - <em>v</em>² / (2<em>a</em>)

Doubling the starting speed gives

- (2<em>v</em>)² / (2<em>a</em>) = - 4<em>v</em>² / (2<em>a</em>) = 4<em>d</em>

so the distance traveled is quadrupled, and it would move a distance of 4 • 15 m = 60 m.

Alternatively, you can explicitly solve for the acceleration, then for the distance:

A car starting at 50 km/h ≈ 13.9 m/s skids to a stop in 15 m, so locked brakes apply an acceleration <em>a</em> such that

0² - (13.9 m/s)² = 2 <em>a</em> (15 m) → <em>a</em> ≈ -6.43 m/s²

So the same car starting at 100 km/h ≈ 27.8 m/s skids to stop over a distance <em>d</em> such that

0² - (27.8 m/s)² = 2 (-6.43 m/s²) <em>d</em> → <em>d</em> ≈ 60 m

(9) Pushing the lever down 1.2 m with a force of 50 N amounts to doing (1.2 m) (50 N) = 60 J of work. So the load on the other end receives 60 J of potential energy. If the acceleration due to gravity is taken to be approximately 10 m/s², then the load has a mass <em>m</em> such that

60 J = <em>m g h</em>

where <em>g</em> = 10 m/s² and <em>h</em> is the height it is lifted, 1.2 m. Solving for <em>m</em> gives

<em>m</em> = (60 J) / ((10 m/s²) (1.2 m)) = 5 kg

(10) Is this also multiple choice? I'm not completely sure, but something about the weight of the tractor seems excessive. It would help to see what the options might be.

4 0
3 years ago
What is the weight of a 4.2 kg bowling ball on Mars?
Nataliya [291]

What is the weight of a 4.2 kg bowling ball on Mars?

Answer:

1.59 kg

Explanation:

The formula is:

<u>F = G((Mm)/r2) </u>

F is the gravitational force between two objects,

G is the Gravitational Constant (6.674×10-11 Newtons x meters2 / kilograms2),

M is the planet's mass (kg),

m is your mass (kg), and

r is the distance (m) between the centers of the two masses (the planet's radius).

Hope this helps

--Jay

8 0
3 years ago
If a charge at 60c flow in a conductor for 30 second then the current that flow in a conductor is​
saw5 [17]

Explanation:

<h3>Given</h3>

- Charge = 60c

time = 30 sec

<h3>To find -</h3>

current

<h3>Solution </h3>

Current = Charge/time

I = V/T

I = 60/30

I = 2 ampere

More to know -

I = Current

V = Charge

T = Time

3 0
3 years ago
5. A box weighs 196 N. A rope is tied to the box. What is the
natta225 [31]

Answer:

296 N

Explanation:

Draw a free body diagram.  The box has two forces on it: tension up and weight down.

Apply Newton's second law:

∑F = ma

T − mg = ma

T = m (g + a)

Given m = 196 N / 9.8 m/s² = 20 kg, and a = +5 m/s²:

T = (20 kg) (9.8 m/s² + 5 m/s²)

T = 296 N

5 0
2 years ago
A flywheel with a very low friction bearing takes 1.6 h to stopafter the motor power
Nina [5.8K]

Answer:

(I). The initial rotation rate is 4.29 rad/s.

(II). The revolutions is 3932.

Explanation:

Given that,

Time = 1.6 h

Angular velocity = 41 rpm

(I). We need to calculate the initial rotation rate in rad/s

\omega=41\ rev/m

\omega=\dfrac{41\times 2\pi}{60}\ rad/s

\omega=4.29\ rad/s

(II). We need to calculate the revolutions

Using formula of revolutions

\theta=\omega t

\theta=4.29\times1.6\times3600

\theta=24710\ rad

\theta=\dfrac{24710}{2\pi}

\theta=3932\ revolution

Hence, (I). The initial rotation rate is 4.293 rad/s.

(II). The revolutions is 3932.

7 0
3 years ago
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