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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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Against the wind a commercial airline in south america flew 784 miles in 4 hours. with a tailwind the return trip took 3.53 hour
ira [324]

First let us assign variables,

d = distance travelled

t = time it took

v = velocity of the commercial airline

In linear physics, the equation for velocity is given as:

v = d / t

Rewriting for d:

d = v t

We know that the distance to and from south America are equal therefore:

d1 (going) = d2 (return)

Let us say that velocity of air is v3. Since going to South America, the wind is against the direction of the plane and the return trip is the opposite, therefore:

(v1 - v3) t1 = (v1 + v3) t2

(v1 – v3) 4 = (v1 + v3) 3.53

4 v1 – 4 v3 = 3.53 v1 + 3.53 v3

0.47 v1 = 7.53 v3

v1 = 16.02 v3

Since we also know that:

(v1 - v3) t1 = 784

(16.02 v3 – v3) * 4 = 784

60.085 v3 = 784

v3 = 13.05 mph

Therefore the speed of the plane in still air, v1 is:

v1 = 16.02 * 13.05

<span>v1 = 209.03 mph           (ANSWER)</span>

<span> </span>

4 0
3 years ago
The maximum current output of a 60 ω circuit is 11 A. What is the rms voltage of the circuit?
bezimeni [28]

Answer:

660V

Explanation:

V=IR

V=11×60

=660V

hope this helps

please like and Mark as brainliest

8 0
3 years ago
If the soccer ball has a mass of 284g, and gravitational force is 9.81 m/s2 on Earth, what is the weight of the soccer ball?
bagirrra123 [75]

Answer:

  1. <u>2.79 N</u>
  2. <u>8.34 N</u>
  3. <u>Watermelon</u>

Explanation:

<u>Weight of soccer ball</u>

  • Weight = mass (in kg) × gravitational force
  • Weight = 284 x 10⁻³ x 9.81
  • Weight = 0.284 x 9.81
  • Weight = <u>2.79 N</u> (approximately)

<u>Weight of watermelon</u>

  • Weight = mass (in kg) x gravitational force
  • Weight = 850 x 10⁻³ x 9.81
  • Weight = 0.85 x 9.81
  • Weight = <u>8.34 N</u> (approximately)

As the <u>watermelon</u> has more weight, it will hit the ground first.

4 0
2 years ago
Technician A says that the cooling system should be tested for leaks using a pressure-operated pressure pump. Technician B says
mart [117]

Answer:

The correct option is;

Technician A only

Explanation:

Leaks in the cooling system parts such as the radiator cap can be tested with a pressure operated hand pump by attaching the pump cap to the pump with an adaptor and apply pressure enough for there to be a release of pressure at the cap.

The pressure added to the system should be maintained for up to two minutes. If the pressure drops quicker, then there is likely to be leaks in the system

The freezing and boiling point of the coolant is measured with an antifreeze tester which carries the appropriate apparatus to measure the coolants freezing and boiling points.

3 0
2 years ago
A track is traveling a. a speed of 25.0 m/s along a level road. A crate is resting on the bed of the truck, and the coefficient
Mice21 [21]

To solve this problem it is necessary to apply the concepts related to

conservation of energy, for this case manifested through work and kinetic energy.

W = \Delta KE

W = F*d

Where,

F= Force (Frictional at this case F_r = \mu N)

d= Distance

\Delta KE = \frac{1}{2} mv^2

Where,

m = mass

v = velocity

Equation both terms,

F*d = \frac{1}{2}mv^2

\mu mg *d = \frac{1}{2}mv^2

\mu g * d = \frac{1}{2}v^2

d = \frac{1}{2} \frac{v^2}{\mu g}

Replacing with our values we have that

d = \frac{1}{2} \frac{25^2}{0.65*9.8}

d = 49.05m

Therefore the shortest distance in which the truck can come to a halt without causing the crate to slip forward relative to the truck is 49.05m

6 0
3 years ago
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