Ask question

Ask question

All categories

3 years ago

11

Calculate the normal force and force applied of a box that has a mass of

1 answer:

GarryVolchara [31]3 years ago

7 0

Heya!!

For calculate force, lets applicate second law of Newton:

$\boxed{F=ma}$

<u>Δ Being Δ</u>

F = Force = ?

m = Mass = 50 kg

a = Aceleration = 7,2 m/s²

⇒ Let's replace according the formula:

$\boxed{F= 50\ kg * 7,2 m/s^{2}}$

⇒ Resolving

$\boxed{ F = 360\ N}$

Result:

The force is <u>360 Netwons.</u>

Good Luck!!

You might be interested in

How do paved parking lots and roads with concrete or asphalt affect water flow on the land?

harina [27]

On the regular ground, water usually seeps in to the soil, which has a number of benefits for the wildlife in that area. However, with asphalt and concrete, there is no soft and absorbent soil to take in the water, so it just keeps flowing down to the lowest part of land it can. Additionally, some water runoff can carry fertilizer and other harmful chemicals with it in to the oceans and lakes it’s dumped in, which harms the ecosystem in them as well.

8 0

3 years ago

Which kind of wave interaction is shown?

Evgesh-ka [11]

Yo my knowledge, refraction.

7 0

3 years ago

Read 2 more answers

One ring of radius a is uniformly charged with charge +Q and is placed so its axis is the x-axis. A second ring with charge –Q i

kati45 [8]

Answer:

The force exerted on an electron is $7.2\times10^{-18}\ N$

Explanation:

Given that,

Charge = 3 μC

Radius a=1 m

Distance = 5 m

We need to calculate the electric field at any point on the axis of a charged ring

Using formula of electric field

$E=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

$E_{1}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times3\times10^{-6}\times5}{(1^2+5^2)^{\frac{3}{2}}}$

$E_{1}=1.0183\times10^{3}\ N/C$

Using formula of electric field again

$E_{2}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times(-3\times10^{-6})\times5}{((0.5)^2+5^2)^{\frac{3}{2}}}$

$E_{2}=-1.064\times10^{3}\ N/C$

We need to calculate the resultant electric field

Using formula of electric field

$E=E_{1}+E_{2}$

Put the value into the formula

$E=1.0183\times10^{3}-1.064\times10^{3}$

$E=-0.045\times10^{3}\ N/C$

We need to calculate the force exerted on an electron

Using formula of electric field

$E = \dfrac{F}{q}$

$F=E\times q$

Put the value into the formula

$F=-0.045\times10^{3}\times(-1.6\times10^{-19})$

$F=7.2\times10^{-18}\ N$

Hence, The force exerted on an electron is $7.2\times10^{-18}\ N$

8 0

3 years ago

Which actions are examples of conserving resources? Check all that apply.

uysha [10]

Answer:

1st, 2nd, and 4th

Explanation:

1st conserves gasoline/petroleum

2nd conserves electricity

4th conserves paper

6 0

3 years ago

Read 2 more answers

A falling skydiver has a mass of 105 kg. What is the magnitude of the skydiver's acceleration when the upward force of air resis

LenKa [72]

F-free = m*g - F_air = m*a

F_air = 1.2 * m

a= (105 kg * 9.8 m.s^2 - 5*105) / 105 kg

a = 9.3 m/s

Hope this helps

7 0

3 years ago