1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Delicious77 [7]
2 years ago
14

The bolts on a car wheel require tightening to a torque of 4 N*m. If a 9 cm long wrench is used, what is the magnitude of the fo

rce required when the force is applied at 8 degrees to the wrench
Physics
1 answer:
notsponge [240]2 years ago
7 0

Answer:

320N

Explanation:

The magnitude of the torque required is expressed using the formula;

T = Fr sin theta where;

F is the force

r is the radius = 9cm = 0.09m

theta is the angle of inclination = 8 degrees

Torque T = 4Nm

Substitute the given values and get F

4 = F(0.09)sin8

4 = 0.0125F

F = 4/0.0125

F = 320N

Hence the magnitude of the force required when the force is applied at 8 degrees to the wrench is 320N

You might be interested in
One day, after pulling down your window shade, you notice that sunlight is passing through a pinhole in the shade and making a s
baherus [9]

Given that,

Central maximum = 1 cm

Distance from the window shade to the wall =4 m

We know that,

The visible range of the sun light is 400 nm to 700 nm.

(a). We need to calculate the average wavelength

Using formula of average wavelength

\lambda_{avg}=\dfrac{\lambda_{1}+\lambda_{2}}{2}

Put the value into the formula

\lambda_{avg}=\dfrac{400+700}{2}

\lambda_{avg}=550\ nm

(b). We need to calculate the diameter of the pinhole

Using formula for diameter

w=\dfrac{2.44\lambda L}{D}

D=\dfrac{2.44\lambda L}{w}

Put the value into the formula

D=\dfrac{2.44\times550\times10^{-9}\times4}{1\times10^{-2}}

D=0.537\ mm

Hence, (a). The average wavelength 550 nm.

(b). The diameter of the pinhole is 0.537 mm.

7 0
3 years ago
When you throw a pebble straight up with initial speed V, it reaches a maximum height H with no air resistance. At what speed sh
aev [14]

Answer:

Explanation:

Given

Initial speed is u=V

Maximum height of Pebble is H

Deriving maximum height of Pebble and considering motion in vertical direction

v^2-u^2=2 as

where v=final velocity

u=initial velocity

a=acceleration

s=Displacement

Final velocity will be zero at maximum height

0-(V)^2=2\times (-g)\cdot H

H=\frac{V^2}{2g}

i.e. maximum height is dependent on square of initial velocity

for twice the height

2H=\frac{(V')^2}{2g}

on comparing

V'=\sqrt{2}V

7 0
3 years ago
A person is standing on an elevator initially at rest at the first floor of a high building. The elevator then begins to ascend
GREYUIT [131]

Answer:

The found acceleration in terms of h and t is:

a=\frac{h}{5(t_1)^2}

Explanation:

(The complete question is given in the attached picture. We need to find the acceleration in terms of h and t in this question)

We are given 3 stages of movement of elevator. We'll first model them each of the stage one by one to find the height covered in each stage. After that we'll find the total height covered by adding heights covered in each stage, and equate it to Total height h. From that we can find the formula for acceleration.

<h3></h3><h3>Stage 1</h3>

Constant acceleration, starts from rest.

Distance = y = \frac{1}{2}a(t_1)^2

Velocity = v_1=at_1

<h3>Stage 2</h3>

Constant velocity where

Velocity = v_o=v_1=at_1

Distance =

<h3>y_2=v_2(t_2)\\\text{Where~}t_2=4t_1 ~\text{and}~ v_2=v_1=at_1\\y_2=(at_1)(4t_1)\\y_2=4a(t_1)^2\\</h3><h3 /><h3>Stage 3</h3>

Constant deceleration where

Velocity = v_0=v_1=at_1

Distance =

y_3=v_1t_3-\frac{1}{2}a(t_3)^2\\\text{Where}~t_3=t_1\\y_3=v_1t_1-\frac{1}{2}a(t_1)^2\\\text{Where}~ v_1t_1=a(t_1)^2\\y_3=a(t_1)^2-\frac{1}{2}a(t_1)^2\\\text{Subtracting both terms:}\\y_3=\frac{1}{2}a(t_1)^2

<h3>Total Height</h3>

Total height = y₁ + y₂ + y₃

Total height = \frac{1}{2}a(t_1)^2+4a(t_1)^2+\frac{1}{2}a(t_1)^2 = 5a(t_1)^2

<h3 /><h3>Acceleration</h3>

Find acceleration by rearranging the found equation of total height.

Total Height = h

h = 5a(t₁)²

a=\frac{h}{5(t_1)^2}

6 0
3 years ago
The magnitude of the gravitational field strength near Earth's surface is represented by
Zanzabum

Answer:

The magnitude of the gravitational field strength near Earth's surface is represented by approximately 9.82\,\frac{m}{s^{2}}.

Explanation:

Let be M and m the masses of the planet and a person standing on the surface of the planet, so that M >> m. The attraction force between the planet and the person is represented by the Newton's Law of Gravitation:

F = G\cdot \frac{M\cdot m}{r^{2}}

Where:

M - Mass of the planet Earth, measured in kilograms.

m - Mass of the person, measured in kilograms.

r - Radius of the Earth, measured in meters.

G - Gravitational constant, measured in \frac{m^{3}}{kg\cdot s^{2}}.

But also, the magnitude of the gravitational field is given by the definition of weight, that is:

F = m \cdot g

Where:

m - Mass of the person, measured in kilograms.

g - Gravity constant, measured in meters per square second.

After comparing this expression with the first one, the following equivalence is found:

g = \frac{G\cdot M}{r^{2}}

Given that G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}, M = 5.972 \times 10^{24}\,kg and r = 6.371 \times 10^{6}\,m, the magnitude of the gravitational field near Earth's surface is:

g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.972\times 10^{24}\,kg)}{(6.371\times 10^{6}\,m)^{2}}

g \approx 9.82\,\frac{m}{s^{2}}

The magnitude of the gravitational field strength near Earth's surface is represented by approximately 9.82\,\frac{m}{s^{2}}.

4 0
2 years ago
A particle is projected with velocity v0 directly up a slope which makes an angle α with the horizontal. Assume frictionless mot
Nikolay [14]

Answer:

a)T total = 2*Voy/(g*sin( α ))

b)α = 0º ,  T total≅∞ (the particle, goes away horizontally indefinitely)

α = 90º,  T total=2*Voy/g

Explanation:

  • Velocity in the Y axis:

Voy=Vo*sinα

  • Time to reach the maximal height :

Kinematics equation: Vfy=Voy-at

a=g*sinα ;  g is gravity

if Vfy=0 ⇒ t=T ; time to reach the maximal height

so:

0=Voy-g*sin( α )*T

T=Voy/(g*sin( α ))

  • Time required to return to the starting point:

After the object reaches its maximum height, the object descends to the starting point, the time it descends is the same as the time it rises.

So T total= 2T = 2*Voy/(g*sin( α ))

  • α = 0º , sinα=0

The particle goes totally horizontal, goes away indefinitely

T total= 2*Voy/(g*sin( α )) ≅∞

  • α = 90º, sinα=1

T total=2*Voy/g

6 0
3 years ago
Other questions:
  • How many nuclear power plants exist in new jersey?
    7·2 answers
  • Brainliest and 100 POINTS
    7·2 answers
  • PLEASE ANSWER THIS WORKSHEET ITS DUE TODAY WILL GIVE BRAINLIEST BALANCED AND UNBALANCED WORKSHEET
    11·1 answer
  • A ball is thrown with an initial speed of 20m/s at an angle of 60 to the ground. If air resistance is negligible, what is the ba
    7·1 answer
  • Since an object moving in uniform circular motion keeps a constant speed, there is no force necessary to keep it in motion.
    15·2 answers
  • A pupil wants to find the density of an oil. She uses a chemical balance which measures to the nearest gram (g). She places an e
    13·1 answer
  • A bicycle rides in a straight line at 1 km at 10
    13·1 answer
  • When an unbalanced force acts on an object
    7·1 answer
  • CAN SOMEONE PLS HELP ME ILL GIVE YOU BRAINLIEST!!!!
    15·2 answers
  • if a girl is running along a straight road with uniform velocity 1.5 metre per second find the acceleration​
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!