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2 years ago

A mover loads a 100 kg box into the back of a moving truck by

Physics

1 answer:

hammer [34]2 years ago

7 0

Explanation:

Mechanical Advantage (MA)

MA=d1d2=FoutFin ; d1 is the distance of effort, d2 is the distance the object is moved

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How does the suns gravitational attraction impact earths motion

Savatey [412]

<span>The lighter object is in orbit around the heaviest, and the Sun is the heaviest object in the Solar System</span>

5 0

2 years ago

An electric light is plugged into a 120-V outlet. If the current in the bulb is 0.50 A, how much electrical energy does the bulb

ioda

Answer:

= 54,000 Joules or 54 kJ

Explanation:

Electrical energy is given by the formula;

E = VIt; where V is the potential difference in volts, I is the current and t is the time in seconds.

Therefore;

Electrical energy = 120 V × 0.50 A × 15 ×60 seconds

= 54,000 Joules

Thus; the electrical energy is 54,000 joules or 54 kJ

7 0

2 years ago

Calculate the Force of Gravity acting on an object that has a mass of 1.3 kg. The object is on Earth so use 9.8m/s2 for the acce

sweet-ann [11.9K]

To calculate the gravitational force acting on an object given the mass and the acceleration due to gravity, use the following formula.

Fg = m • g
Fg = 1.3 kg • 9.8 m/s^2
Fg = 12.74 N or about 12.7 N.

The solution is C. 12.7 N.

7 0

2 years ago

An object weighing 49 N is pushed across a floor by a force of 12 N. What is the acceleration of the object?

NISA [10]

Answer:

Explanation:

Given parameters:

Weight of object = 49N

Force applied = 12N

Unknown:

Acceleration of object = ?

Solution:

The acceleration of the object is found by dividing the force by the weight;

Acceleration = $\frac{12}{49}$ = 0.25m/s²

3 0

2 years ago

A simple harmonic oscillator has amplitude 0.43 m and period 3.9 sec. What is the maximum acceleration?

rjkz [21]

Answer:

Maximum acceleration in the simple harmonic motion will be $0.854rad/sec^2$

Explanation:

We have given amplitude of simple harmonic motion is A = 0.43 m

Time period of the oscillation is T = 3.9 sec

We have to find the maximum acceleration

For this we have to find the angular frequency

Angular frequency will be equal to $\omega =\frac{2\pi }{T}=\frac{2\times 3.14}{3.9}=1.61rad/sec$

Maximum acceleration is given by $a_{max}=\omega ^2A=1.61^2\times 0.43=0.854rad/sec^2$

So maximum acceleration in the simple harmonic motion will be $0.854rad/sec^2$

8 0

3 years ago

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