All categories

3 years ago

A man pushes on a wall what does Newton third law say must happen

2 answers:

8 0

If a man pushes on a wall with some force then according to Newton's third law, wall will also apply force on man with same magnitude but opposite in direction.

aivan3 [116]3 years ago

3 0

The man will receive an equal force back, because for every action there is an equal and opposite reaction

You might be interested in

You weigh 716 newtons on Earth. You

Otrada [13]

Answer:

537 N

Explanation:

The force due to gravity of a planet is:

F = GMm / r²

where G is the universal gravitational constant

M is the mass of the planet

m is the mass of the object

and r is the distance between the object and the center of the planet

On Earth, you weigh 716 N, so:

716 N = GMm / r²

On planet X:

F = G (3M) m / (2r)²

F = 3/4 GMm / r²

F = 3/4 (716 N)

F = 537 N

8 0

4 years ago

a 5.0 kg ball is dropped from a 2.5 m high window. what is the velocity of the ball just before it hits the ground?

julsineya [31]

Answer:

Approximately $7.0\; \rm m \cdot s^{-1}$ .

Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.

Explanation:

Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant $g = 9.81 \; \rm m \cdot s^{-2}$ near the surface of the earth.

For an object that is accelerating constantly,

$v^2 - u^2 = 2\, a \, x$ ,

where

$u$ is the initial velocity of the object,
$v$ is the final velocity of the object.
$a$ is its acceleration, and
$x$ is its displacement.

In this case, $x$ is the same as the change in the ball's height: $x = 2.5\; \rm m$ . By assumption, this ball was dropped with no initial velocity. As a result, $u = 0$ . Since the ball is accelerating due to gravity, $a = 9.81\; \rm m \cdot s^{-2}$ .

$v^2 - u^2 = 2\, g \cdot h$ .

In this case, $v$ would be the velocity of the ball just before it hits the ground. Solve for

$v^2 = 2\, a\, x + u^2$ .

$\begin{aligned}v &= \sqrt{2\, a\, x + u^2} \\ &= \sqrt{2\times 9.81 \times 2.5 + 0} \\ &\approx 7.0\; \rm m\cdot s^{-1}\end{aligned}$ .

3 0

3 years ago

a canoe is floating down a river with a velocity of 35 meters per minute when it suddenly approaches a waterfall. if the waterfa

vampirchik [111]

From Newton's law v^2 = u^2 + 2as where a is the acceleration and s is the distance.
But to go any further, we need to know how fast the vehicle is accelerating
From v = u +at
We have a = u/t where the final velocity v = 0

So in one minute acceleration = (35 / 60) / 60 = 0.0097 ms/2. The first
experession in bracket is the initial velocity, u, in metres per seconds.
Hence v^2 = (0.583)^2 + 2 (0.0097)(30)
v^2 = 0.3398 + 0.5826 = 0.9224
v = âš 0.9224 = 0.960m

3 0

3 years ago

A box weights 50kg applies friction force of 125N. You apply a force of 150N going in the right direction. What is the total net

Pachacha [2.7K]

Answer:

Net force = 25N and in the right direction

Explanation:

mass of box, m = 50kg

Force of block = 150N

Friction force = 125N

$Net force = Force Applied - Friction force = F - F_f$

150 - 125 = 25N

the force applied to the block is greater than the friction force. there the force applied will overcome the friction force and move in the right direction of the force applied

4 0

4 years ago

A particle with a charge of 2e moves between two points which have a potential difference of 75V. What is the change in potentia

Sonbull [250]

Electric potential energy is defined as Ep=Q*V where Q is the magnitude of the charge and V is the potential difference. So when a charge moves between the points that have a potential difference, it's energy changes.

In our case:

Q=2e=2*(-1.6*10^-19) C
V=75 V

Ep=(-3.2*10^-19)*75

Ep=-2.4*10^-17 J

The change in potential energy of the charge is -2.4*10^-17 J

5 0

3 years ago