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
12345 [234]
3 years ago
9

The normal force of a parked car is 15,000 Newtons. The coefficient of static friction between the rubber of the tires and the a

sphalt of the road is 0.75. What is the maximum static friction force?
Physics
1 answer:
Shtirlitz [24]3 years ago
3 0

Answer:

11250 N

Explanation:

From the question given above, the following data were obtained:

Normal force (R) = 15000 N

Coefficient of static friction (μ) = 0.75

Frictional force (F) =?

Friction and normal force are related by the following equation:

F = μR

Where:

F is the frictional force.

μ is the coefficient of static friction.

R is the normal force.

With the above formula, we can calculate the frictional force acting on the car as follow:

Normal force (R) = 15000 N

Coefficient of static friction (μ) = 0.75

Frictional force (F) =?

F = μR

F = 0.75 × 15000

F = 11250 N

Therefore, the frictional force acting on the car is 11250 N

You might be interested in
CAN YOU HELP PLZZ & THX
maxonik [38]

Answer:

6.7x10 -7 C

Explanation:

The -7 is like a exponent on top, I think.

6 0
3 years ago
what is change in internal energy if 30j of heat are released from a system that does 50j of work on its surroundings
slamgirl [31]

Answer:

-80 J

Explanation:

The first law of thermodynamics states that:

\Delta U = Q - W

where

\Delta U is the change in internal energy of the system

Q is the heat absorbed by the system

W is the work done by the system

In this problem, we have:

Q = -30 J is the heat released by the system (negative because the system releases it)

W = +50 J is the work done by the system on the surrounding (positive since it is done by the system)

Therefore, the change in internal energy is

\Delta U = -30 J - (+50 J) = -80 J

3 0
4 years ago
A girl with a mass of 27 kg is playing on a swing. There are three main forces
N76 [4]

The tension in the swing's chain at the bottom of the swing is 178.35 N.

The given parameters:

  • Mass of the girl, m = 27 kg
  • Speed of the girl, v = 3 m/s
  • Radius of the circle, r = 4 m

The tension in the swing's chain at the bottom of the swing is calculated as follows;

T = mg + ma_c\\\\ T= mg + \frac{mv^2}{r} \\\\T = (12 \times 9.8) + (\frac{27 \times 3^2}{4} )\\\\T = 117.6 \ N \ + \ 60.75 \ N\\\\T = 178.35 \ N

Thus, the tension in the swing's chain at the bottom of the swing is 178.35 N.

Learn more about tension in vertical circle here: brainly.com/question/19904705

3 0
2 years ago
If the force is 4 newtons between two charged spheres separated by 3 centimeters, calculate the force between the same spheres s
il63 [147K]

Answer:

1 Newton

Explanation:

F=9*10^9*q0q1/r^2]]

F=9*10^9*(q0q1)/ r^2

r=3cm

F=4N

F=9*10^9*(q0q1)/3^2

4=9*10^9*(q0q1)/9

4=10^9 q0q1

q0q1=4/10^9

q0q1=4*10^-9

To calculate the force between the forces at a distance of 6 cm

F=9*10^9*(q0q1)/ r^2

=9*10^9*(4*10^-9)/6^2

=9*10^9*(4*10^-9)/36

=10^9*4*10^-9/4

=10^9*10^-9

=1 Newton

7 0
3 years ago
a particle that has a mass of 2.5 kg is moving in the positive X-direction with a constant velocity of 1.6m/s. Suddenly a consta
malfutka [58]
The particle has a constant horizontal velocity, and a vertical force won't affect the horizontal speed, so it should be fairly easy to find the last part, "the time taken for a 10m horizontal displacement," using a kinematic equation.
X = x + vt + (1/2)at²
10 = 0 + (1.6)t + (1/2)(0)t²
10/1.6 = t
t = 6.25s

So now we have to find the vertical displacement over 6.25 seconds on a particle of a 2.5kg mass with a force of 8N.
Start with Newton's second law.
F = ma
8 = (2.5)a
a = 3.2m/s²

Now, use kinematics again.
Y = y + vt + (1/2)at²
Y = 0 + (0)(6.25) + (1/2)(3.2)(6.25)²
Y = <u>62.5m</u>
7 0
3 years ago
Other questions:
  • At a certain point between Earth and the Moon, the net gravitational force exerted on an object by Earth and the Moon is zero. T
    10·1 answer
  • Trade winds are found ________.
    15·1 answer
  • Two solutions, initially at 24.60°C, are mixed in a coffee cup calorimeter (Ccal = 15.5 J/°
    15·1 answer
  • Melissa and Rory live near a bike trail, 21 km apart. They plan to meet at a park mid-way between their homes. They start from t
    10·1 answer
  • What are some other examples of a force acting on an object without<br> touching it?
    10·1 answer
  • An object, with mass 72 kg and speed 28 m/s relative to an observer, explodes into two pieces, one 2 times as massive as the oth
    9·1 answer
  • Can sound travel through space? Why or why not?
    10·1 answer
  • Will give brainliest!!
    15·1 answer
  • A baseball is hit with a speed of 27.0 m/s at an angle of 47.0 ∘ . It lands on the flat roof of a 10.0 m -tall nearby building.
    12·1 answer
  • What is Kinetic Energy and Potential energy❓<br><br>Ty!!​
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!