Ask question

Ask question

All categories

3 years ago

15

A chair of weight 125 N lies atop a horizontal floor; the floor is not frictionless. You push on the chair with a force of F = 3

5.0 N directed at an angle of 42.0 degrees below the horizontal and the chair slides along the floor.
Using Newton's laws, calculate n, the magnitude of the normal force that the floor exerts on the chair.

1 answer:

iris [78.8K]3 years ago

5 0

Answer:

N = 148.10N

Explanation:

GIVEN

Weight =125 N

force = 35

angle =42°

since there is no vertical acceleration $a_{y} = 0$ from

free body diagram

$\Sigma fa_{y} = ma_{y} = 0$

$N-W-Fsin42\degree = 0\\\\N= W+Fsin42\\N = 125+35\times 0.66\\N = 148.10 N$

You might be interested in

this event is the result of a sudden vertical offset in the ocean floor and is most often the result of plate movement on the oc

MissTica

Answer: Tsunami.

Explanation :

Tsunami is the result of a sudden vertical offset in the ocean floor and is most often the result of plate movement on the ocean floor.

Tsunami is caused due to the displacement of a large volume of water like in an ocean. It consists of a series of waves. It destroys coastlines and coastal settlements. It is also known as a tidal wave.

So, the correct option is (b) Tsunami.

6 0

3 years ago

The boiling point of water at sea level is 100 °c. at higher altitudes, the boiling point of water will be

Scilla [17]

<span> The boiling point of water at sea level is 100 °C. At higher altitudes, the boiling point of water will be.....
a) higher, because the altitude is greater.
b) lower, because temperatures are lower.
c) the same, because water always boils at 100 °C.
d) higher, because there are fewer water molecules in the air.
==> e) lower, because the atmospheric pressure is lower.
--------------------------
Water boils at a lower temperature on top of a mountain because there is less air pressure on the molecules.
-------------------
I hope this is helpful. </span>

8 0

3 years ago

The length of a simple pendulum is 0.81 mand the mass of the particle (the "bob") at the end of the cable is0.23 kg. The pendulu

Gemiola [76]

Answer:

$\displaystyle w=3.478\ rad/sec$

$M=0.0182\ J$

$v=0.398\ m/s$

Explanation:

<u>Simple Pendulum</u>

It's a simple device constructed with a mass (bob) tied to the end of an inextensible rope of length L and let swing back and forth at small angles. The movement is referred to as Simple Harmonic Motion (SHM).

(a) The angular frequency of the motion is computed as

$\displaystyle w=\sqrt{\frac{g}{L}}$

We have the length of the pendulum is L=0.81 meters, then we have

$\displaystyle w=\sqrt{\frac{9.8}{0.81}}$

$\displaystyle w=3.478\ rad/sec$

(b) The total mechanical energy is computed as the sum of the kinetic energy K and the potential energy U. At its highest point, the kinetic energy is zero, so the mechanical energy is pure potential energy, which is computed as

$U=mgh$

where h is measured to the reference level (the lowest point). Please check the figure below, to see the desired height is denoted as Y. We know that

$H+Y=L$

And

$H=L\ cos\alpha$

Solving for Y

$Y=L(1-cos\alpha )$

$Since\ \alpha=8.1^o, L=0.81\ m$

$Y=0.0081\ m$

The potential energy is

$U=mgh=0.23\ kg(9.8\ m/s^2)(0.0081\ m)$

$U=0.0182\ J$

The mechanical energy is, then

$M=K+U=0+U=U$

$M=0.0182\ J$

(c) The maximum speed is achieved when it passes through the lowest point (the reference for h=0), so the mechanical energy becomes all kinetic energy (K). We know

$\displaystyle K=\frac{mv^2}{2}$

Equating to the mechanical energy of the system (M)

$\displaystyle \frac{mv^2}{2}=0.0182$

Solving for v

$\displaystyle v=\sqrt{\frac{(2)(0.0182)}{0.23}}$

$v=0.398\ m/s$

4 0

3 years ago

How much potential energy does a 40n block medicine ball gain when it is lifted 5.m

MAVERICK [17]

By definition, the gain in PE (potential energy) is
ΔPE = m*g*h

Given:
mg = 40 N (Note that m*g = weight)
h = 5 m

ΔPE = (40 N)*(5 m) =200 J

Answer: 200 J

5 0

3 years ago

Read 2 more answers

A 2.5 kg tribble is placed in a bucket and whirled in a 1.4 m radius vertical circle at a constant tangential speed. If the forc

Over [174]

Given that,

Mass of a tribble, m = 2.5 kg

Radius, r = 1.4 m

The force on the tribble from the bucket does not exceed 10 times its weight.

To find,

The maximum tangential speed.

Solution,

The force acting on the tribble is equal to the centripetal force.

F = 10mg

The formula for the centripetal force is given by :

$F=\dfrac{mv^2}{r}$

v is maximum tangential speed

$v=\sqrt{\dfrac{Fr}{m}} \\\\v=\sqrt{\dfrac{mgr}{m}} \\\\v=\sqrt{{10gr}} \\\\v=\sqrt{10\times 9.8\times 1.4} \\\\v=11.7\ m/s$

So, the maximum tangential speed is 11.7 m/s.

8 0

3 years ago