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Ne4ueva [31]
2 years ago
11

A Mercedes-Benz 300SL (m = 1700 kg) is parked on a road that rises 15 degrees above the horizontal. What are the magnitudes of (

a) the normal force and (b) the static frictional force that the ground exerts on the tires?
Physics
1 answer:
kogti [31]2 years ago
4 0

Answer: See below

Explanation:

<u>Given:</u>

Mass of the Mercedes-Benz (m) = 1700 kg

Inclination of the road (θ) = 15.0

<em>The free body diagram is shown in figure attached below</em>

<em />

a) The normal force is equal to the cos component of the weight of the car.

\begin{aligned}&f=m g \cos \theta \\&f=1700 \times 9.81 \times \cos 15 \\&f=16108.74 \mathrm{~N}\end{aligned}

b) The static force will be equal to the weight's sin component.

\begin{aligned}&f=m g \sin \theta \\&f=1700 \times 9.81 \times \sin 15 \\&f=4316.32 \mathrm{~N}\end{aligned}

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A cannonball is fired at a 45.0° angle and an initial velocity of 670 m/s. Assume no air resistance. What is the vertical compon
elixir [45]

Answer:

<h3>473.8 m/s; 473.8 m/s</h3>

Explanation:

Given the initial velocity U = 670m/s

Horizontal velocity Ux = Ucos theta

Vertical component of the cannon velocity Uy = Usin theta

Given

U = 670m/s

theta = 45°

horizontal component of the cannonball’s velocity = 670 cos 45

horizontal component of the cannonball’s velocity = 670(0.7071)

horizontal component of the cannonball’s velocity = 473.757m/s

Vertical component of the cannonball’s velocity = 670 sin 45

Vertical component of the cannonball’s velocity  = 670 (0.7071)

Vertical component of the cannonball’s velocity  = 473.757m/s

Hence pair of answer is 473.8 m/s; 473.8 m/s

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The site of protein synthesis is
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3 years ago
Stress distributed over an area is best described as: a) External force b) Axial force c) Radial force d) Internal resistive for
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Answer:

Option D is the correct answer.

Explanation:

Stress is the force per unit area that tend to change the shape of body.

Stress is defined as internal resistive force per unit area.

         \texttt{Stress}=\frac{\texttt{Internal resistive force}}{\texttt{Area}}

         \sigma =\frac{F}{A}

So, so stress distributed over an area is best described as internal resistive force.

Option D is the correct answer.

8 0
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In a particular case of an object in front of a spherical mirror with a focal length of +12.0 cm, the magnification is +4.00.(a)
salantis [7]

Answer:

9 cm

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Explanation:

u = Object distance

v = Image distance

f = Focal length = 12

m = Magnification = 4

m=-\frac{v}{u}\\\Rightarrow 4=-\frac{v}{u}\\\Rightarrow v=-4u

Lens equation

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{12}=\frac{1}{u}+\frac{1}{-4u}\\\Rightarrow \frac{1}{12}=\frac{3}{4u}\\\Rightarrow u=9\ cm

Object distance is 9 cm

v=-4\times 9=-36\ cm

Image distance is -36 cm (other side of object)

7 0
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