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m_a_m_a [10]
3 years ago
7

Find the net force and acceleration. 10 points. Will give brainliest.

Physics
2 answers:
Svetllana [295]3 years ago
5 0

\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}

Actually Welcome to the Concept of the FRICTION.

Force by Xavier = 25 N

Force of friction in the opposite direction of Xavier = f = umg

where, u = coefficient of kinetic friction.

so we here get as ,

Net force = 16 N

and acceleration = 2.13 m/s^2

Afina-wow [57]3 years ago
3 0

Answer:

a) 16 N

b) 2.13 m/s²

Explanation:

Draw a free body diagram of the tv stand.  There are four forces:

Weight force mg pulling down,

Normal force N pushing up,

Friction force Nμ pushing left,

and applied force P pulling right.

Sum of forces in the y direction:

∑F = ma

N − mg = 0

N = mg

The net force in the x direction is:

∑F = P − Nμ

∑F = P − mgμ

∑F = 25 N − (7.5 kg) (10 m/s²) (0.12)

∑F = 16 N

Net force equals mass times acceleration:

∑F = ma

16 N = (7.5 kg) a

a = 2.13 m/s²

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A meter stick is held vertically with one end on the floor and is then allowed to fall. Find the speed of the other end when it
Tems11 [23]

Answer:

5.4 ms⁻¹

Explanation:

Here we have to use conservation of energy. Initially when the stick is held vertical, its center of mass is at some height above the ground, hence the stick has some gravitational potential energy. As the stick is allowed to fall, its rotates about one. gravitational potential energy of the stick gets converted into rotational kinetic energy.

L = length of the meter stick = 1 m

m = mass of the meter stick

w = angular speed of the meter stick as it hits the floor

v = speed of the other end of the stick

we know that, linear speed and angular speed are related as

v = r w\\w = \frac{v}{r}

h = height of center of mass of meter stick above the floor = \frac{L}{2} = \frac{1}{2} = 0.5 m

I = Moment of inertia of the stick about one end

For a stick, momentof inertia about one end has the formula as

I = \frac{mL^{2} }{3}

Using conservation of energy

Rotational kinetic energy of the stick = gravitational potential energy

(0.5) I w^{2} = mgh\\(0.5)(\frac{mL^{2} }{3}) (\frac{v}{L} )^{2} = mgh\\(0.5)(\frac{v^{2} }{3}) = gh\\(0.5)(\frac{v^{2} }{3}) = (9.8)(0.5)\\v = 5.4 ms^{-1}

7 0
3 years ago
When an object experiences uniform circular motion the direction of the net force is?
guajiro [1.7K]
In an uniform circular motion, the direction of the net force on the object is radially inward, passing through the center of the circle.
3 0
3 years ago
Nuclear fusion is when two atoms of __________________ join together to form _____________.
solong [7]

Answer:

1. Hydrogen
2. Helium

Explanation:

Nuclear fusion is when two atoms of Hydrogen join together to form one Helium atom.

3 0
2 years ago
3. A football is kicked with a speed of 35 m/s at an angle of 40°.
jarptica [38.1K]

a) 22.5 m/s

The initial vertical velocity is given by:

u_y = u sin \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_y = (35)(sin 40)=22.5 m/s

b) 26.8 m/s

The initial horizontal velocity is given by:

u_x = u cos \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_x = (35)(cos 40)=26.8 m/s

c) 2.30 s

The time it takes for the ball to reach the maximum heigth can be found by considering the vertical motion only. This is a uniformly accelerated motion (free-fall), so we can use the suvat equation

v_y = u_y + at

where

v_y is the vertical velocity at time t

u_y = 22.5 m/s

a=g=-9.8 m/s^2 is the acceleration of gravity (negative because it is downward)

At the maximum height, the vertical velocity becomes zero, v_y =0; substituting, we find the time t at which this happens:

0=u_y + gt\\t=-\frac{u_y}{g}=-\frac{22.5}{-9.8}=2.30 s

d) 25.8 m

The maximum height can also be found by considering the vertical motion only. We can use the following suvat equation:

s=u_y t + \frac{1}{2}gt^2

where

s is the vertical displacement at time t

u_y = 22.5 m/s

g=-9.8 m/s^2

Substituting t = 2.30 s, we find the displacement at maximum height, so the maximum height:

s=(22.5)(2.30)+\frac{1}{2}(-9.8)(2.30)^2=25.8 m

e) 123.3 m

In order to find how far does the ball lands, we have to consider the horizontal motion.

First of all, the time it takes for the ball to go back to the ground is twice the time needed for reaching the maximum height:

t=2(2.30 s)=4.60 s

Then, we consider the horizontal motion. There is no acceleration along this direction, so the horizontal velocity is constant:

v_x = 26.8 m/s

Therefore, the horizontal distance travelled during the whole motion is

d=v_x t = (26.8)(4.60)=123.3 m

So, the ball lands 123.3 m far from the initial point.

4 0
3 years ago
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Ilia_Sergeevich [38]

Answer:

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Explanation:

7 0
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