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3 years ago

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Objects 1 and 2 attract each other with a gravitational force of 72.0 units. If the distance separating Objects 1 and 2 is chang

ed to one-third the original value, then the new gravitational force will be units

1 answer:

gizmo_the_mogwai [7]3 years ago

3 0

Answer:

It is 24

Explanation:

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An elevator is pulled up by a cable with a force of 65,000 N. The upward acceleration of the elevator is 1.8 m/s/s. What is the

goblinko [34]

Answer:

36111 kg

Explanation:

Given

Force = 65000N

Acceleration = 1.8m/s²

Required

Determine the mass of the elevator

This question will be answered using the following Force formula.

Force = Mass * Acceleration

Substitute values for Force and Acceleration

65000N = Mass * 1.8m/s²

Make Mass the subject

Mass = 65000N/1.8m/s²

Mass = 36111.11 kg

From the list of given options, option E answers the question.

6 0

3 years ago

An alternating source drives a series RLC circuit with an emf amplitude of 6.34 V, at a phase angle of +33.3°. When the potentia

Katena32 [7]

Answer:

V $_{L}$ =- 8.78v

Explanation:

$E^{2} _{m}$ = $V^{2} _{R}$ + $(V_{L} -V _{C} )^{2}$

OR V $_{L}$ = V $_{c}$ + $\sqrt{E^{2}{m} - V^{2} {R} }$

where VR = E ${m}$ cosФ

V $_{L}$ = V $_{c}$ + $\sqrt{E^{2}_{m} - E^{2}_{m} cos^{2} }$ Ф

V $_{L}$ = V $_{c}$ + ${E_{m} \sqrt{1 - cos^{2} } }$ Ф

V $_{L}$ = V $_{c}$ + E $_{m}$ sinФ

substituting the given values

V $_{L}$ = 5.33 + 6.34 x sin33°

V $_{L}$ = -8.78v

8 0

3 years ago

Read 2 more answers

if a bicyclist, with initial velocity of zero, steadily gain speed until reaching a final velocity of 26m/s, how far away did sh

Alja [10]

Answer:

<h2><em>34.46m</em></h2>

Explanation:

Using one of the equation of motion to solve the question. According to the equation v² =u²+2as where;

v is the final velocity of the bicyclist = 26m/s

u is the initial velocity of the bicyclist = 0m/s

a is the acceleration due to gravity = 9.81m/s

s is the distance covered during travel

Substitute the given parameters into the formula above to get the distance traveled

26² = 0² + 2(9.81)s

676 = 19.62s

Divide both sides by 19.62

676/19.62 = 19.62s/19.62

s = 34.46m

<em>The distance traveled by the bicyclist during the race is 34.46m</em>

5 0

3 years ago

A 0.700-kg particle has a speed of 1.90 m/s at point circled A and kinetic energy of 7.20 J at point circled B. (a) What is its

Savatey [412]

Answer:

a). $E_{kA}=1.2635 J$

b). $V_{B}=4.535\frac{m}{s}$

c). Δ $E_{t}=8.4635 J$

Explanation:

ΔE=kinetic energy

a).

$E_{kA}=\frac{1}{2}*m*v_{A} ^{2} \\ v_{A}=1.9 \frac{m}{s}\\ m=0.70kg\\E_{kA}=\frac{1}{2}*0.70kg*(1.9 \frac{m}{s})^{2} \\E_{kA}=1.2635 J$

b).

$E_{kB}=\frac{1}{2}*m*v_{B} ^{2}$

$V_{B}^{2}=\frac{E_{kB}*2}{m} \\V_{B}=\sqrt{\frac{E_{kB}*2}{m}} \\V_{B}=\sqrt{\frac{7.2J*2}{0.70kg}} \\V_{B}=4.53 \frac{m}{s}$

c).

net work= EkA+EkB

$E_{t}=E_{kA}+ E_{kB}\\E_{t}=1.2635J+7.2J\\E_{t}=8.4635J$

3 0

3 years ago

A planet orbits a start with the path shown below.

kvv77 [185]

PART a)

As we know that gravitational potential energy is given by the formula

$U = -\frac{Gm_1m_2}{r}$

here we can see that gravitational potential energy inversely varies with the distance

so here when distance from the sun is minimum then magnitude of gravitational potential energy is maximum while since it is given with negative sign so its overall value is minimum at that position

So gravitational potential energy is minimum at the nearest point and maximum at the farthest point

PART b)

Since we know that sum of kinetic energy and potential energy is constant here

so the points of minimum potential energy is the point where kinetic energy is maximum which means speed is maximum

So here speed is maximum at the nearest point

Part C)

since gravitational potential energy inversely varies with distance so it's graph will be like hyperbolic graph with distance

4 0

3 years ago