Ask question

Ask question

All categories

3 years ago

10

8. An object has a force applied. The object has a mass of 36 Kg and accelerates at 3.4 m/sec2. What is the force acting on the

object? *
a. 120.6 N
b. 126.9 N
c. 122.4 N
d. 118.3 N

1 answer:

satela [25.4K]3 years ago

7 0

Answer:

122.4

Explanation:

You might be interested in

Reply quick pls

Sholpan [36]

The ball was moving for 0.8seconds
This is because 4/5 x 1 = 0.8

Hope this helps and good luckkkk :)

4 0

2 years ago

A cold glass is left sitting outside on a hot day. Soon, water droplets form on the outside of the glass. Describe the events th

ikadub [295]

Evaporation must happen

8 0

3 years ago

A4 40 kg girl skates at 3.5 m/s one ice toward her 65 kg friend who is standing still, with open arms. As they collide and hold

salantis [7]

Explanation:

Below is an attachment containing the solution.

8 0

3 years ago

2. Suppose that a parallel-plate capacitor has circular plates with radius R = 30 mm and a plate separation of d = 5.0 mm. Suppo

seropon [69]

Answer:

The maximum value of the induced magnetic field is $2.901\times10^{-13}\ T$ .

Explanation:

Given that,

Radius of plate = 30 mm

Separation = 5.0 mm

Frequency = 60 Hz

Suppose the maximum potential difference is 100 V and r= 130 mm.

We need to calculate the angular frequency

Using formula of angular frequency

$\omega=2\pi f$

Put the value into the formula

$\omega=2\times\pi\times60$

$\omega=376.9\ rad/s$

When r>R, the magnetic field is inversely proportional to the r.

We need to calculate the maximum value of the induced magnetic field that occurs at r = R

Using formula of magnetic filed

$B_{max}=\dfrac{\mu_{0}\epsilon_{0}R^2\timesV_{max}\times\omega}{2rd}$

Where, R = radius of plate

d = plate separation

V = voltage

Put the value into the formula

$B_{max}=\dfrac{4\pi\times10^{-7}\times8.85\times10^{-12}\times(30\times10^{-3})^2\times100\times376.9}{2\times130\times10^{-3}\times5.0\times10^{-3}}$

$B_{max}=2.901\times10^{-13}\ T$

Hence, The maximum value of the induced magnetic field is $2.901\times10^{-13}\ T$ .

7 0

3 years ago

Stellar nurseries, such as the orion nebula, contain hundreds or more fragmenting and contracting regions, as well as many proto

Natasha_Volkova [10]

The hydrogen fusion process will begin after the protostar reaches a temperature of 10 million degrees kelvin, and it will then turn into a stable star.

<h3>How does a protostar become a stable star?</h3>

The interstellar medium can sometimes be gathered into a large nebula, which is a cloud of gas and dust. A nebula can span a number of light years. These nebulae are where gas and dust can combine to produce stars. Until a star can combine hydrogen into helium, it cannot be considered a star. They are referred to as protostars before then. As gravity starts to gather the gases into a ball, a protostar is created. Accrution is the term for this procedure.

Gravitational energy starts to heat the gasses as gravity draws them into the ball's core, which causes the gasses to radiate radiation. Radiation initially just dissipates into space. However, much of the radiation is retained inside the protostar as it draws in stuff and becomes denser, which causes the protostar to heat up even more quickly.

The hydrogen fusion process will begin after the protostar reaches a temperature of 10 million degrees kelvin, and it will then turn into a star.

Learn more about a protostar here:

brainly.com/question/12534975

#SPJ4

3 0

1 year ago