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3 years ago

If all else stays the same, which would cause an increase in the gravitational force on a space shuttle?

Physics

2 answers:

Alenkasestr [34]3 years ago

8 0

Answer: Decreasing the distance of the space shuttle from Earth .

Explanation:

According to expression of gravitational force:

$F=\frac{G\times m_1\times m_2}{r^2}$

G = gravitational constant

$m_1, m_2$ = masses of two objects

r = Distance between the two objects.

F = Gravitational force

From the above expression we can say that gravitational force is inversely proportional to squared of the distance between the two masses.

$F\propto \frac{1}{r^2}$

So, in order to increase the gravitational force on space shuttle distance between the space space shuttle must be decreased.

Hence, the correct answer 'decreasing the distance of the space shuttle from Earth '.

Troyanec [42]3 years ago

8 0

Answer:

The correct answer is B.

Explanation:

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Distinguishing between Speed and Velocity.

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Answer:

Speed: Distance per time, 400 km/h, and a scalar quantity.

Velocity: Displacement per time, 20 m/s south, and a vector quantity.

Explanation:

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2 years ago

Magnetic fields go from the north poles of magnets to the south poles of magnets.<br> true or false

Novay_Z [31]

The Answer is true, the saying opposites attract are true for magnets when poles match though they repel.

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3 years ago

A mechanic changing the spark plugs in a car notes that the instruction manual calls for a torque with a magnitude of

Papessa [141]

The magnitude (in N) of the force she must exert on the wrench is 150.1 N.

<h3>Force exerted by the wrench</h3>

The force exerted by the wrench is calculated using torque formula as follows;

torque, τ = F x r x sinθ

where;

F is the applied force
r is the perpendicular distance if force applied

F = τ /(r sinθ)

F = (39) / (0.3 sin 60)

F = 150.1 N

Thus, the magnitude (in N) of the force she must exert on the wrench is 150.1 N.

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1 year ago

Suppose a police officer is 1/2 mile south of an intersection, driving north towards the intersection at 40 mph. At the same tim

blagie [28]

Answer:

75.36 mph

Explanation:

The distance between the other car and the intersection is,

$x=x_{0}+V t \\ x=\frac{1}{2}+V t$

The distance between the police car and the intersection is,

$y=y_{0}+V t$

$y=\frac{1}{2}-40 t$

(Negative sign indicates that he is moving towards the intersection)

Therefore the distance between them is given by,

$z^{2}=x^{2}+y^{2}(\text { Using Phythogorous theorem })$

$z^{2}=\left(\frac{1}{2}+V t\right)^{2}+\left(\frac{1}{2}-40 t\right)^{2} \ldots \ldots \ldots(1)$

The rate of change is,

$2 z \frac{d z}{d t}=2\left(\frac{1}{2}+V t\right) V+2\left(\frac{1}{2}-40 t\right)(-40)$

$2 z \frac{d z}{d t}=V+2 V^{2} t-40+3200 t \ldots \ldots \ldots$

Now finding $z$ when $t=0,$ from (1) we have

$z^{2}=\left(\frac{1}{2}+V(0)\right)^{2}+\left(\frac{1}{2}-40(0)\right)^{2}$

$z^{2}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2} \\ z=\sqrt{\frac{1}{2}} \approx 0.7071$

The officer's radar gun indicates 25 mph pointed at the other car then, $\frac{d z}{d t}=25$ when $t=0,$ from

From (2) we get

$2(0.7071)(25)=V+2 V^{2}(0)-40+3200(0)$

$2(0.7071)(25)=V+2 V^{2}(0)-40$

$35.36=V-40$

$V=35.36+40=75.36$

Hence the speed of the car is $75.36 mph$

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3 years ago

4 properties of light

Neko [114]

-reflection and refraction of light
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-absorption of light
-polarization of light

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2 years ago