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2 years ago

13

A skater moves with 15 m/s velocity in a circle of radius circle of radius 30 m. The ice exerts a center force of 450 N. What is

the mass of the skater?

1 answer:

wariber [46]2 years ago

6 0

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Amy wants to know whether or not an item will float when placed in a fluid. Which of the comparisons below, when true, will mean

o-na [289]

A. The object is made of stone or metal, which always sink.

B. The force of gravity is stronger than the buoyant force.

C. The buoyant force is stronger than the force of gravity.

D. The mass of the item is greater than its gravity

The answer is C

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2 years ago

Which type of seismic wave can only pass through the earths mantle?

mezya [45]

Secondary wave is the answer

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3 years ago

A person weighing 0.9 kN rides in an elevator that has a downward acceleration of 1.9 m/s^2. The acceleration of gravity is 9.8

Keith_Richards [23]

Answer:

The magnitude of the force is 0.7255kN

Explanation:

The elevator floor acts on the person with a force that is due to the gravitational acceleration less the downward acceleration of the elevator:

(force of floor F) = (mass of person m) x [ (grav. acceleration g) - (elevator acceleration a) ]

in other words, considering the elevator floor as a reference frame in the Earth's gravitational field, the person's weight decreases due to the downward acceleration, as follows:

$F = m\cdot(g-a)$

We are given the person's weight at rest, 0.9kN, from which the mass can be determined as:

$900 N = m\cdot g \implies m = \frac{900N}{9.8 \frac{m}{s^2}}$

So

$F = \frac{900N}{9.8 \frac{m}{s^2}}\cdot(9.8-1.9)\frac{m}{s^2}\approx 725.5N=0.7255kN$

3 0

2 years ago

How does a transverse wave move

Yanka [14]

They move in a waves motion

6 0

3 years ago

Read 2 more answers

A car starts from rest and accelerates uniformly at 2.0 m/s2 toward the north. A second car starts from rest 4.0 s later at the

yKpoI14uk [10]

Answer:

the correct solution is 13 s

Explanation:

This is a kinematic problem, let's use accelerated rectilinear motion relationships.

For the first car it has an accelerometer of 2.0 m/s²

x = v₀₁ t + ½ a₁ t²

The second car leaves the same point, but 4.0 seconds later

x = v₀₂ (t-4) + ½ a₂ (t-4)²

With this form we use the same time for both cars.

The initial speeds are zero for both vehicles leave the rest, at the point where they are located has the same position

x = ½ a₁ t²

x = ½ a₂ (t-4)²

Let's solve

a₁ t² = a₂ (t-4)²

a₁/a₂ t² = t² -2 4 t + 16

t² (1- 2.0 / 4.0) - 8 t +16

t² 0.5 - 8 t +16 = 0

t² -16 t + 32 = 0

Let's solve the second degree equation

t = [16 ±√( 16² - 4 32)] / 2

t = ½ (16 ± 11,3)

Solutions

t1 = 13.66 s

t2 = 2.34 s

These are the mathematical solutions for the meeting point, but car 2 leaves after 4 seconds, so the only solution is 13.66 s

the correct solution is 13 s, if you have to select one the nearest 12s

6 0

3 years ago