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3 years ago

What are 3 things that can affect gravity?

1 answer:

5 0

The force of gravity the masses exert on each other. If one of the masses is doubled , the force of gravity between the objects is doubled. Increases , the force of gravity decreases.

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A gull is flying horizontally 10.80 m above the ground at 6.00 m/s. The bird is carrying a clam in its beak and plans to crack t

UkoKoshka [18]

Answer:

v_y = 14.55 m/s

Explanation:

given,

height at which gull is flying = 10.80 m

speed of the gull = 6 m/s

acceleration due to gravity = 9.8 m/s²

Relative to the seagull, the x-speed is 0,

because the seagull has the same x-speed.

Only the y-speed counts:

$v_y^2 = u^2 + 2 g h$

$v_y^2 = 0^2 + 2 g h$

$v_y = \sqrt{2gh}$

$v_y = \sqrt(2\times 9.8 \times 10.8)$

$v_y = \sqrt(211.68)$

v_y = 14.55 m/s

hence, the speed at which the clam smash the rock is v_y = 14.55 m/s

8 0

3 years ago

A force in the +x-direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 8.90 kg box that is sitting on the horizontal,

Lady_Fox [76]

Answer:

6.875 m/s

Explanation:

The force is variable which is given by

F(x) = 18 - 0.53 x

mass of the box, m = 8.9 kg

initially it is at rest at x = 0

Let the velocity is v after travelling a distance of 15 m.

According to the work energy theorem, the work done by all the forces is equal to the change in kinetic energy of the body.

Work done = change in kinetic energy

$\int \overrightarrow{F}.d\overrightarrow{x}=\Delta K.E$

$\int_{0}^{15} \left ( 18-0.53 x \right )dx=\frac{1}{2}\times m \left ( v^{2}-u^{2} \right )$

$\left ( 18x-0.265x^{2} \right )_{0}^{15}=\frac{1}{2}\times 8.9\times \left ( v^{2}-0^{2} \right )$

18 x 15 - 0.265 x 15 x 15 = 4.45 x v²

270 - 59.625 = 4.45 v²

v² = 47.275

v = 6.875 m/s

Thus, the final velocity of the box is 6.875 m/s.

4 0

3 years ago

For the graph above, what is the independent variable? What is the dependant variable?

posledela

Time/sec. the x-axis is always the independent variable.

3 0

3 years ago

What is the average speed of a car that travels 60 meters in 2<br> seconds?

blondinia [14]

Answer:

30 m/s

Explanation:

Speed is distance over time. 60 meters / 2 seconds, = 30 m/s.

6 0

3 years ago

What average net force is required to stop a 3.5 kg bowling ball initially travelling at a speed of 1.5 m/s over a distance of 0

raketka [301]

First the aceleration:

Vf² = Vo² - 2ad

a = (Vf² - Vo²) / 2d

a = (0 m/s)² - (1,5 m/s)²) / 2 * 0,4 m

a = -2,25 m²/s² / 0,8 m

a = -2,81 m/s²

Now, for the net force, use 2nd law of Newton:

F = ma

F = 3,5 kg * (-2,81 m/s²)

F = -9,835 N

The force for stop the bowling ball is <u>-9,835 Newtons.</u>

7 0

3 years ago