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nataly862011 [7]
3 years ago
13

When the spacecraft is at the halfway point, how does the strength of the gravitional force on the spaceprobe by Earth compre wi

th the strength
Physics
1 answer:
mixer [17]3 years ago
5 0

Solution :

When the spacecraft is at halfway point, the distance from the Earth as well as Mars are same. We have to account the masses of the planets. The gravitational force that is exerted by the Earth is greater because of its combined mass with the space probe.

The mass of Earth is greater than the mass of Mars. Therefore, the force of Earth is more than Mars.

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A bat strikes a 0.145-kg baseball. Just before impact, the ball is traveling horizontally to the right at 60.0 m/s , and it leav
nata0808 [166]

Answer:

F = -307.4 N

Explanation:

It is given that,

Mass of the baseball, m = 0.145 kg

Initial speed of the baseball, u = 60 m/s

Final speed of the baseball, v=65\ cos(30)=56.29\ m/s

Time of contact, t=1.75\ ms=1.75\times 10^{-3}\ s

(a) It is assumed to find the horizontal component of average force. It is given by :

F=m\dfrac{v-u}{t}

F=0.145\dfrac{56.29-60}{1.75\times 10^{-3}}

F = -307.4 N

So, the horizontal component of average force is 307.4 N. Hence, this is the required solution.

8 0
3 years ago
A hot air balloon is traveling vertically upward at a constant speed of 4.5 m/s. When it is 28 m above the ground, a package is
ella [17]

Answer:

1.97 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

s=ut+\frac{1}{2}at^2\\\Rightarrow 21=4.5t+\frac{1}{2}\times -9.8\times t^2\\\Rightarrow 21=-4.5t-4.9t^2\\\Rightarrow 4.9t^2+4.5t-28=0\\\Rightarrow 49t^2+45t-280=0

Solving the above equation we get

t=\frac{-45+\sqrt{45^2-4\cdot \:49\left(-280\right)}}{2\cdot \:49}, \frac{-45-\sqrt{45^2-4\cdot \:49\left(-280\right)}}{2\cdot \:49}\\\Rightarrow t=1.97, -2.89

So, the time the package was in the air is 1.97 seconds

3 0
3 years ago
A block weighing 10 newtons is resting on a plane inclined 30° to the horizontal. What is the magnitude of the normal force acti
aleksley [76]

Magnitude of normal force acting on the block is 7 N

Explanation:

10N = 1.02kg

Mass of the block = m = 1.02 kg

Angle of incline Θ =  30°

Normal force acting on the block = N

From the free body diagram,

N = mgCos Θ

N = (1.02)(9.81)Cos(30)

N = 8.66 N

Rounding off to nearest whole number,

N = 7 N

Magnitude of normal force acting on the block = 7 N

7 0
3 years ago
classic physics problem states that if a projectile is shot vertically up into the air with an initial velocity of 128 feet per
ddd [48]

Answer:

t_1 = 0.28 s

t_2 = 7.72 s

Explanation:

Given that height of the projectile as a function of time is

h = -16 t^2 + 128 t + 112

here we know that

h = 147 ft

so from above equation

147 = -16 t^2 + 128 t + 112

16 t^2 - 128 t + 35 = 0

now by solving above quadratic equation we know that

t_1 = 0.28 s

t_2 = 7.72 s

6 0
3 years ago
An artificial satellite in a low orbit circles the earth every 98 minutes. what is its angular speed in rad/s?
postnew [5]
To finish one orbit it will take 98 x 60 seconds. So; <span>(2 x pi)/(98 x 60) = 1.07 x 10^-3 rad/sec. </span><span>
</span>
8 0
3 years ago
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