How much time would the weight take to fall 120 m down the cliff if it was thrown downwards at 1.25 m/s?

A 0.500-kg mass suspended from a spring oscillates with a period of 1.50 s. How much mass must be added to the object to change

Daniel [21]

Answer:

389 kg

Explanation:

The computation of mass is shown below:-

$T = 2\pi \sqrt{\frac{m}{k} }$

Where K indicates spring constant

m indicates mass

For the new time period

$T^' = 2\pi \sqrt{\frac{m'}{k} }$

Now, we will take 2 ratios of the time period

$\frac{T}{T'} = \sqrt{\frac{m}{m'} }$

$\frac{1.50}{2.00} = \sqrt{\frac{0.500}{m'} }$

$0.5625 = \sqrt{\frac{0.500}{m'} }$

$m' = \frac{0.500}{0.5625}$

= 0.889 kg

Since mass to be sum that is

= 0.889 - 0.500

0.389 kg

or

= 389 kg

Therefore for computing the mass we simply applied the above formula.

7 0

3 years ago

A coil has an inductance of 6.00 mH, and the current in it changes from 0.200 A to 1.50 A in a time interval of 0.300 s. Find th

Andre45 [30]

Answer:

0.026 V

Explanation:

Given that,

Inductance of the coil, L = 6 mH

The current changes from 0.2 A to 1.5 A in a time interval of 0.3 s

We need to find the magnitude of the average induced emf in the coil during this time interval. The formula for the induced emf is given by :

$\epsilon=L\dfrac{dI}{dt}\\\\=6\times 10^{-3}\times \dfrac{1.5-0.2}{0.3}\\\\=0.026\ V$

So, the magnitude of induced emf is 0.026 volts.

5 0

3 years ago

Two particles with charges of 2nC and 5nC are separated by a distance of 3m. The charge 2nC is placed on the left. Find the forc

Vanyuwa [196]

Answer:

$1\cdot 10^{-8} N$ to the left

Explanation:

The magnitude of the electrostatic force between two charges is given by the following equation:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=9\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the magnitude of the two charges

r is the distance between the two charges

Moreover, the force is:

- Attractive if the charges have opposite sign

- Repulsive if the charges have same sign

In this problem, we have:

$q_1=2nC=2\cdot 10^{-9}C$ is the magnitude of charge 1

$q_2=5nC =5\cdot 10^{-9}C$ is the magnitude of charge 2

r = 3 m is the distance between the two charges

Substituting, we find the force on both charges:

$F=\frac{(9\cdot 10^9)(5\cdot 10^{-9})(2\cdot 10^{-9}}{3^2}=1\cdot 10^{-8} N$

Here, the two charges are both positive, so the force is repulsive; since the 2 nC charge is on the left, this means that the force on this charge is to the left (away from the 5 nC charge).

5 0

3 years ago

What is the SI (metric) unit of FORCE? A. meter B. newton

Tresset [83]

What is the SI (metric) unit of FORCE?

B. newton

with symbol ( N )

All the best !

4 0

3 years ago

Read 2 more answers

The velocity of a 640-kg auto is changed from 10.0 m/s to 44.0 m/s in 71.0 s by an external, constant force. (a) What is the res

n200080 [17]

Answer:

(a) change in momentum of the car = 21760 kgm/s

(b) The magnitude of the force = 306.48 N

Explanation:

(a)

Change in momentum: This is the product of mass and change in velocity. The unit of momentum is kgm/s. It can be expressed mathematically as,

Change in momentum of the car = m(v-u)................. Equation 1

Where m = mass of the car, v = final velocity of the car, u = initial velocity of the car.

Given: m = 640 kg, v = 44.0 m/s, u = 10.0 m/s.

Substituting these values into equation 1,

Change in momentum of the car = 640 (44-10)

change in momentum of the car = 640× 34

change in momentum of the car = 21760 kgm/s

change in momentum of the car = 21760 kgm/s

(b) Force: Force of a body can be defined as the product of mass and its acceleration. It is measured in Newton (N). It can be expressed mathematically as,

Force = ma ................................ Equation 2

 Or

Force = m(v-u)/t...................................... Equation 3

Where m = mass of the car, v = final velocity of the car, u = initial velocity of the car, a = acceleration of the car, t = time taken to change the velocity.

Given: m = 640 kg, u = 10.0 m/s v = 44 m/s, t = 71 s.

Substituting these values into equation 3

Force = 640(44-10)/71

Force = (640×34)/71

Force = 306.48 N

The magnitude of the force = 306.48 N

3 0

3 years ago