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4 years ago

Protons and neutrons in an atom are held together by _____.

Physics

1 answer:

marysya [2.9K]4 years ago

7 0

Protons and neutrons in an atom are held together by a nuclear energy also called the strong force.

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according to newtons third law of motion what happened to the back of a skateboard when the person pushes down on the front of t

WINSTONCH [101]

Answer:I think it would start moving

Explanation:

3 0

3 years ago

suppose we have two masses m1=2000 g and m2=4000g, where m1 is moving with initial velocity v1,i=24m/s and m2 is at rest at t=0s

Veseljchak [2.6K]

Answer:

The final velocity is 8 m/s and its direction is along the positive x-axis

Explanation:

Given :

Mass, m₁ = 2000 g = 2 kg

Mass, m₂ = 4000 g = 4 kg

Initial velocity of mass m₁, v₁ = 24i m/s

Initial velocity of mass m₂, v₂ = 0

According to the problem, after collision the two masses are stick together and moving with same velocity, that is, $v_{1f}$ .

Applying conservation of momentum,

Momentum before collision = Momentum after collision

$m_{1} v_{1} +m_{2} v_{2} =(m_{1}+m_{2}) v_{1f}$

Substitute the suitable values in the above equation.

$2\times24 +4\times0 =(2+4}) v_{1f}$

$v_{1f}=8i\ m/s$

8 0

3 years ago

Atmospheric pressure decreases with increment in height.<br> give reason

jekas [21]

Answer:

pressure=height × density×acc due to gravity

pressire is directly proportional to height hence it decreses with decrease in height

here air column height is measured upside down so decreases witn increment

7 0

3 years ago

Read 2 more answers

What is the net torque on the square plate, with sides 0.2 m, from each of the three forces? F1=18 N, F2=26 N, and F3=14 N. Use

marshall27 [118]

Answer:

The net torque on the square plate is 2.72 N-m.

Explanation:

Given that,

Side = 0.2 m

Force $F_{1}=18\ N$

Force $F_{2}=26\ N$

Force $F_{3}=14\ N$

We need to calculate the torque due to force F₁

Using formula of torque

$\tau_{1}=-F_{1}d_{1}$

$\tau_{1}=-F_{1}\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{1}=-18\times\dfrac{0.2}{2}$

$\tau_{1}=-1.8\ N-m$

We need to calculate the torque due to force F₂

Using formula of torque

$\tau_{2}=F_{2}d_{2}$

$\tau_{2}=F_{2}\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{2}=26\times\dfrac{0.2}{2}$

$\tau_{2}=2.6\ N-m$

We need to calculate the torque due to force F₃

Using formula of torque

$\tau_{3}=F_{3}d_{3}$

$\tau_{3}=(F_{3}\sin45+F_{3}\cos45)\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{3}=0.1(14\sin45+14\cos45)$

$\tau_{3}=1.92\ N-m$

We need to calculate the net torque on the square plate

$\tau=\tau_{1}+\tau_{2}+\tau_{3}$

$\tau=-1.8+2.6+1.92$

$\tau=2.72\ N-m$

Hence, The net torque on the square plate is 2.72 N-m.

3 0

4 years ago

A force of 7.50 N is applied to a spring whose spring constant is .298 N/cm. Find it’s length

gizmo_the_mogwai [7]

Answer:

to find the length just divide 7.50 and . 298

5 0

3 years ago