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5. The force of gravity acting on the object is object’s ______________.

Dima020 [189]

I think it's mass...

The mass of an object refers to the amount of matter that is contained by the object; the weight of an object is the force of gravity acting upon that object. Mass is related to how much stuff is there and weight is related to the pull of the Earth (or any other planet) upon that stuff.

5 0

3 years ago

At what distance will a 45 W lightbulb have the same apparent brightness as a 130 W bulb viewed from a distance of 35 m ? (Assum

Sloan [31]

Answer:

20.6 m

Explanation:

P₁ = Power of first bulb = 45 W

P₂ = Power of second bulb = 130 W

r₁ = distance from the first bulb

r₂ = distance from the second bulb = 35 m

Using the equation

$\frac{P_{1}}{r_{1}^{2}} = \frac{P_{2}}{r_{2}^{2}}$

Inserting the values

$\frac{45}{r_{1}^{2}} = \frac{130}{35^{2}}$

r₁ = 20.6 m

5 0

4 years ago

A mass is undergoing simple harmonic motion. When its displacement is 0, it is at its equilibrium position. At that moment, its

hram777 [196]

Answer:

The speed is maximum and the acceleration is zero

Explanation:

- The speed of the mass in simple harmonic motion can be found by using the law of conservation of energy. In fact, the total mechanical energy of the mass-spring system is sum of kinetic energy and elastic potential energy:

$E=K+U=\frac{1}{2}mv^2+\frac{1}{2}kx^2$

where

m is the mass

v is the speed

k is the spring constant

x is the displacement

As we can see, when the displacement is zero (x=0), the term representing the kinetic energy is maximum, so v (the speed) is also maximum.

- The acceleration of the mass in simple harmonic motion is proportional to the restoring force acting on the mass, which is given by Hook's law

$a \propto F = -kx$

where

k is the spring constant

x is the displacement

When x = 0, F = 0, so the net force acting on the mass is zero. Therefore, this also means that the acceleration of the mass is also zero: a = 0.

3 0

3 years ago

Read 2 more answers

A rocket is launched upward with a constant acceleration of 165 m/s^2. After 8.00 seconds of ascension a passenger on the rocket

Novay_Z [31]

To solve this problem we will apply the concepts related to the linear kinematic movement. We will start by finding the speed of the body from time and the acceleration given.

Through the position equations we will calculate the distance traveled.

Finally, using this same position relationship and considering the previously found speed, we can determine the time to reach your goal.

For time (t) and acceleration (a) we have to,

$t = 8s, a = 165m/s^2$

The velocity would be,

$u = a*t \\u = 165*8\\u = 1320m/s$

Now the position is,

$h= \frac{1}{2} at^2$

$h = \frac{1}{2} 165*8^2$

$h = 5280m$

Now with the initial speed and position found we will have the time is,

$h=ut +\frac{1}{2} at^2$

$-5280=1320t - \frac{1}{2} 9.8t^2$

$4.9t^2-1320t-5280=0$

Solving the polynomian we have,

$t = 273.33s = 4.56minutes$

Therefore the rocket will take to hit the ground around to 4.56min

5 0

3 years ago

A ball falls for 9.0 s increasing its kinetic energy by 2700. If the force acting on the ball in 6.0 N find the following quanti

Ivanshal [37]

Answer:

Change in velocity: 88 m/s

Average velocity: 50 m/s

initial velocity: 5.9 m/s

Final velocity: 94 m/s

Initial momentum: 3.6 kg m/s

Final momentum: 58 kg m/s

Explanation:

Acceleration = change in velocity / time

9.8 m/s² = Δv / 9.0 s

Δv = 88 m/s

Work = change in energy

Fd = ΔE

(6.0 N) d = 2700 J

d = 450 m

Average velocity = distance / time

v_avg = 450 m / 9.0 s

v_avg = 50 m/s

v − v₀ = 88 m/s

½ (v + v₀) = 50 m/s

Solving the system of equations:

v + v₀ = 100 m/s

2v = 188 m/s

v = 94 m/s

v₀ = 5.9 m/s

Use Newton's second law to find the mass:

F = ma

6.0 N = m (9.8 m/s²)

m = 0.61 kg

Find the momentums:

p₀ = (0.61 kg) (5.9 m/s) = 3.6 kg m/s

p = (0.61 kg) (94 m/s) = 58 kg m/s

6 0

3 years ago