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Bond [772]
3 years ago
13

Is the normal force equivalent to the weight of an object?

Physics
1 answer:
Greeley [361]3 years ago
4 0

Answer:

No.

Explanation:

If there are no applied forces, normal force is usually equivalent to the weight of the object but if there are outside force (force that makes the object to move) especially if it's inclined, then the inclined force would then affect the normal force.

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A hair dryer is a pump for air. Do a battery and a Genecon act like pumps for charge?
lyudmila [28]

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No

Explanation:

8 0
3 years ago
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Identify the following as
bazaltina [42]
3. Kinetic energy
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6 0
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A net force of 200 newtons is applied to a wagon for 3 seconds. This cause the wagons to undergo a change in momentum of ?
suter [353]
Physics- damon, Monday, December 1, 2014 at 3:27 pm force =change in momentum\ change in time or m a if m is constant 

change in momentum/3=200

change in momentum =3*200 kg m/s





8 0
3 years ago
Define moment of momentum. at which condition is it's magnitude zero?​
ololo11 [35]

Let's start with the concept of momentum. What is it? Linear momentum in physics is mathematically written as a product of mass and velocity of an object. Now let us suppose a body of mass m is moving in an inertial frame of reference with velocity v. Consider the fact that no external force is acting on the system. The momentum of this body is given by mv, where m is the mass and v is its velocity. In case of simple real world problems not delving into the realms of relativity, mass is a conserved quantity and it cannot be zero. Hence the velocity of the body must be zero and hence the momentum.

However, photons are considered to have a rest mass zero.

However note the point carefully "rest mass". A body in motion cannot have mass to be zero.

<em>-</em><em> </em><em>BRAINLIEST</em><em> answerer</em><em> ❤️</em>

7 0
3 years ago
Guys please help me ​
Likurg_2 [28]

Answer:

1)t=2.26\: s

2)S=33.9\: m

3)v=26.77\: m/s

4)\alpha=55.92

Explanation:

1)

We can use the following equation:

y_{f}=y_{0}+v_{iy}t-0.5*g*t^{2}

Here, the initial velocity in the y-direction is zero, the final y position is zero and the initial y position is 25 m.

0=25-0.5*9.81*t^{2}

t=2.26\: s

2)

The equation of the motion in the x-direction is:

v_{ix}=\frac{S}{t}

15=\frac{S}{2.26}

S=33.9\: m

3)

The velocity in the y-direction of the stone will be:

v_{fy}=v_{iy}-gt

v_{fy}=0-(9.81*2.26)

v_{fy}=-22.17\: m/s

Now, the velocity in the x-direction is 15 m/s then the velocity will be:

v=\sqrt{v_{x}^{2}+v_{fy}^{2}}=\sqrt{15^{2}+(-22.17)^{2}}

v=26.77\: m/s

4)

The angle of this velocity is:

tan(\alpha)=\frac{22.17}{15}

\alpha=tan^{-1}(\frac{22.17}{15})

\alpha=55.92

Then α=55.92° negative from the x-direction.

I hope it helps you!

6 0
3 years ago
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