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3 years ago

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The Sun being the largest body has the highest gravitational pull on all planets however planets do not collapse into the sun th

is is because of A.gravitational I'll of the Earth B. gravitational pull among the planets C. gravitational pull of Moon D. the balancing force

1 answer:

mojhsa [17]3 years ago

7 0

Answer:

A:gravitational of the earth

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3 years ago

A ball is thrown with velocity of 10 m/s upwards. If the ball is caught 1 m above its initial position, what is the speed of the

White raven [17]

Answer:

v = 8.96 m/s

Explanation:

Initial speed of the ball, u = 10 m/s

It caught 1 meter above its initial position.

Acceleration due to gravity, $g=-9.8\ m/s^2$

We need to find the final speed of the ball when it is caught. Let is equal to v. To find the value of v, use third equation of motion as :

$v^2-u^2=2as$

$v^2=2as+u^2$

$v^2=2(-9.8)\times 1+(10)^2$

v = 8.96 m/s

So, the speed of the ball when it is caught is 8.96 m/s. Hence, this is the required solution.

6 0

3 years ago

Read 2 more answers

At 20°c, the resistance of a sample of nickel is 525 ohms. what is the resistance when the sample is heated to 70°C?

Ipatiy [6.2K]

The resistance of the sample is $682.5\Omega$

Explanation:

The relationship between resistance of a material and temperature is given by the equation

$R(T)=R_0(1+\alpha (T-T_0))$

where

$R_0$ is the resistance at the temperature $T_0$

$\alpha$ is the temperature coefficient of resistance

For the sample of nickel in this problem, we have:

$R_0 = 525 \Omega$ when the temperature is $T_0 = 20^{\circ}C$

While the temperature coefficient of resistance of nickel is

$\alpha = 0.006/^{\circ}C$

Therefore, the resistance of the sample when its temperature is

$T=70^{\circ}C$

is

$R=(525)(1+0.006(70-20))=682.5 \Omega$

Learn more about resistance:

brainly.com/question/4438943

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3 years ago

Before a collision, a 50.0-kg object is moving at +5.0 m/s. Find the impulse that acted on the object if, after the collision, i

Virty [35]

Answer:

<em>J=600 kg m/s </em>

Explanation:

<u>Impulse And Momentum </u>

Suppose a particle is moving at a certain speed $v_1$ and changes it to $v_2$ . The impulse J is equivalent to the change of linear momentum. The momentum can be computed by

$p=mv$

The initial and final momentums are given, respectively, by:

$p_1=mv_1,\ p_2=mv_2$

Thus, the change of momentum is

$\Delta p=p_2-p_1=m(v_2-v_1)$

It's equal to the Impulse J

$J=\Delta p$

$J=m(v_2-v_1)$

Our data is

$m=50\ kg,\ v_1=5\ m/s,\ v_2=17\ m/s$

$J=50\ (17-5)$

$J=600\ kg\ m/s$

7 0

3 years ago

The escape velocity of a bullet from the surface of planet Y is 1695.0 m/s. Calculate the escape velocity from the surface of th

serg [7]

The escape velocity from the surface of the planet X is 2,249.2 m/s.

<h3>Escape velocity of planet X</h3>

$v = \sqrt{\frac{2GM}{r} } \\\\v^2 = \frac{2GM}{r}\\\\v^2r = 2GM\\\\G = \frac{v^2r}{2M}$

where;

M is mass of the planet
r is radius of the planet
G is universal gravitation constant

$\frac{v_x^2 \ r_x}{2M_x} = \frac{v_y^2 \ r_y}{2M_y} \\\\\frac{v_x^2 \ r_x}{M_x} = \frac{v_y^2 \ r_y}{M_y} \\\\v_x^2 = \frac{v_y^2 \ r_yM_x}{M_yr_x}\\\\v_x^2 = \frac{(1695)^2 (r_y)(1.59M_y)}{M_y(0.903r_y)} \\\\v_x^2 = 5,058,814.78\\\\v_x = \sqrt{5,058,814.78} \ \ = 2,249.2 \ m/s$

Thus, the escape velocity from the surface of the planet X is 2,249.2 m/s.

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2 years ago