All categories

3 years ago

What type of path does a moving object follow in the absence of a force?.

Physics

1 answer:

Amanda [17]3 years ago

3 0

Answer:

When there is no force acting on a body then a moving object move in a straight line with constant velocity and an object in rest stays in rest.

Explanation:

When there is no force acting on a body then a moving object move in a straight line with constant velocity and an object in rest stays in rest.

You might be interested in

An LC circuit consists of a 3.14 mH inductor and a 5.08 µF capacitor. (a) Find its impedance at 55.7 Hz. 563.57 Correct: Your an

ANEK [815]

Answer:

$z=561.7$

$z=214.1$

Explanation:

L = inductance of the Inductor = 3.14 mH = 0.00314 H

C = capacitance of the capacitor = 5.08 x 10⁻⁶ F

f = frequency = 55.7 Hz

Impedance is given as

$z=\frac{1}{2\pi fC} - 2\pi fL$

$z=\frac{1}{2(3.14) (55.7)(5.08\times 10^{-6})} - 2(3.14) (55.7)(0.00314)$

$z=561.7$

f = frequency = 11000 Hz

Impedance is given as

$z= - \frac{1}{2\pi fC} + 2\pi fL$

$z= - \frac{1}{2(3.14) (11000)(5.08\times 10^{-6})} + 2(3.14) (11000)(0.00314)$

$z=214.1$

4 0

3 years ago

4. Two forces act on a 2 kg object as shown. What is the magnitude of the acceleration of the object?

aleksandrvk [35]

The resultant force on the object is

∑ F = 〈0, 8〉 N + 〈6, 0〉 N = 〈6, 8〉 N

which has a magnitude of

F = √((6 N)² + (8 N)²) = √(100 N²) = 10 N

By Newton's second law, the acceleration has magnitude a such that

F = m a

10 N = (2 kg) a

a = (10 N) / (2 kg)

a = 5 m/s²

so the answer is B.

4 0

3 years ago

Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150

Sever21 [200]

Answer: $2.89\approx 2.9^{\circ}C/s$

Explanation:

Given

$Power\left ( P\right )=150 MW$

mass of core $\left ( m\right )=1.60\times 10^5 kg$

Average specific heat $\left ( C\right )=0.3349 KJ/kg^{\circ}C$

And rate of increase of temperature = $\frac{\mathrm{d}T}{\mathrm{d} t}$

Now

P= $mc\frac{\mathrm{d}T}{\mathrm{d} t}$

$150\times 10^6=1.60\times 10^5\times 0.3349\times \frac{\mathrm{d}T}{\mathrm{d} t}$

Thus $\frac{\mathrm{d}T}{\mathrm{d} t}=[tex]\frac{1.60\times 10^5\times 0.3349}{150\times 10^6}$

$\frac{\mathrm{d}T}{\mathrm{d} t}=2.89\approx 2.9^{\circ}C/s$

6 0

3 years ago

A dart gun consists of a horizontal spring with k = 52 Newtons/m that is compressed 43

gulaghasi [49]

Answer:

$299 m/s^2$

Explanation:

When a spring is compressed, the force exerted by the spring is given by:

$F=kx$

where

k is the spring constant

x is the compression of the spring

In this problem we have:

k = 52 N/m is the spring constant

x = 43 cm = 0.43 m is the compression

Therefore, the force exerted by the spring on the dart is

$F=(52)(0.43)=22.4 N$

Now we can apply Newton' second law of motion to calculate the acceleration of the dart:

$F=ma$

where

F = 22.4 N is the force exerted on the dart by the spring

m = 75 g = 0.075 kg is the mass of the dart

a is its acceleration

Solving for a,

$a=\frac{F}{m}=\frac{22.4}{0.075}=299 m/s^2$

7 0

3 years ago

A piece of rope is pulled by two people in a tug-of-war. each exerts a 400-n force. what is the tension in the rope?

Leokris [45]

Newtons 3.law: Action = Reaction

If a body exerts a force on a rope of 400 N the rope exerts a force on the body of 400N also. So the tension in the rope is 400N. See pictures below.

3 0

3 years ago

Read 2 more answers