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4 years ago

Which force acts as an object to move it from rest or a constant straight-line motion?

Physics

1 answer:

podryga [215]4 years ago

8 0

Answer: C

Explanation: The correct answer would be Inertial. I take the quiz and got the answer rght!!

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You took a running leap off a high diving platform. You were running at 3.1 m/s and hit the water 2.4 seconds later. How high wa

rusak2 [61]

The distance you free-fall from rest is D = (1/2) (g) (T²) <== memorize this

Height of the platform = (1/2) (9.8 m/s²) (2.4 sec)²

Height = (4.9 m/s²) (5.76 s²)

Height = (4.9/5.76) meters

Height = 28.2 meters (a VERY high platform ... about 93 ft off the water !)

Without air-resistance, your horizontal speed doesn't change. It's constant. Traveling 3.1 m/s for 2.4 sec, you cover (3.1 m/s x 2/4 s) = 7.4 m horizontally.

7 0

4 years ago

Which of the following objects is accelerating?

jenyasd209 [6]

Answer:

Explanation:

Constant speed (without change in direction) is not accelerating. If you are slowing down, speeding up, or changing direction, you are accelerating

4 0

3 years ago

A 300-n force acts on a 25-kg object. the acceleration of the object is?

Over [174]

To solve the answer use the equation: a = fnet / m

a = 300 N / 25 kg

300 N / 25 kg = 12m/s

The acceleration of the object is 12m/s

8 0

3 years ago

Read 2 more answers

Joe drove at the speed of 45 miles per hour for a certain distance. He then drove at the speed of 55 miles per hour for the same

Snowcat [4.5K]

Answer:

$v_{avg} = 49.5 mph$

Explanation:

Let the distance moved by Joe is "d"

so the time taken by him to drove it by speed 45 mph is given as

$t_1 = \frac{d}{v_1}$

$t_1 = \frac{d}{45}$

now the same distance is traveled by him with speed 55 mph

so the time taken by him

$t_2 = \frac{d}{55}$

so total time taken by him for complete distance 2d

$t = t_1 + t_2$

$t = \frac{d}{45} + \frac{d}{55}$

$t = 0.0404 d$

now the average speed is given as

$v_{avg} = \frac{2d}{t}$

$v_{avg} = \frac{2d}{0.0404d}$

$v_{avg} = 49.5 mph$

5 0

3 years ago

Read 2 more answers

The cylinder in the picture is rotating at 500 RPMs. The friction coefficients between the cylinder and block B are static=0.5 a

Tju [1.3M]

Coefficient of static friction = 0.5
Coefficient of Kinetic friction = 0.3
Angular velocity = 500 RPMs

<h3>The Radius of the System</h3>

Let R be the radius of cylinder

$m_a + m_b = 4 + 3 = 7kg$

The angular velocity is 500 RPMs

$\omega ^2 = \frac{500 * 2\pi}{60} rad/s\\N = (M_a + M_b)\omega ^2 R$

The normal force

$f = \mu N = (M_a + M_b) g\\\mu (M_a + M_b) \omega ^2 R = M_a + M_b\\R = \frac{1}{\mu \omega ^2 R}\\\mu_s = 0.5\\R = \frac{1}{0.5 * (\frac{500 * 2\pi}{60})^2 }\\R = 0.0073m\\R = 7.3mm$

Since the radius is very little for two block to execute circular motion so system will slide down.

Learn more on coefficient of static friction here;

brainly.com/question/11841776

brainly.com/question/25772665

brainly.com/question/26400616

8 0

2 years ago