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dolphi86 [110]
3 years ago
5

A car advertisement states that a certain car can accelerate from rest to

Physics
2 answers:
kaheart [24]3 years ago
6 0

Answer:

a = 2.22 [m/s^2]

Explanation:

First we have to convert from kilometers per hour to meters per second

40 [\frac{km}{h}]*[\frac{1h}{3600s}]*[\frac{1000m}{1km}] = 11.11 [m/s]

We have to use the following kinematics equation:

v_{f}= v_{i}+(a*t)

where:

Vf = final velocity = 11.11 [m/s]

Vi = initial velocity = 0

a = acceleration [m/s^2]

t = time = 5 [s]

The initial speed is taken as zero, as the car starts from zero.

11.11 = 0 + (a*5)

a = 2.22 [m/s^2]

Aliun [14]3 years ago
5 0

Refer to the attachment.

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If a car takes a banked curve at less than a given speed, friction is needed to keep it from sliding toward the inside of the cu
zubka84 [21]

Answer:

minimum speed is 15.35 m/s

frictional coefficient  is 0.26

Explanation:

given data

radius = 84 m

angle = 16°

speed = 16 km/h = 4.43 m/s

to find out

minimum speed and   minimum coefficient

solution

we will apply here formula for velocity that is

velocity² = radius × g × tanθ

v² = 84 × 9.8 × tan16

v² =  236.04

v = 15.35 m/s

and

we find first friction force here

friction force 1 = m v² /r

friction force 1 = m (15.35)² / 84 = 2.80 m

and

friction force 2 = m v² /r

friction force 2 = m (4.43)² / 84 =  0.245 m

so total friction force = f1 - f2

total friction force = 2.80 - 0.245  = 2.55 m

so frictional coefficient = friction force /g

frictional coefficient = 2.55 / 9.8

so frictional coefficient  is 0.26

4 0
3 years ago
Read 2 more answers
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
3 years ago
Charge X has twice as much charge as particle Y. The two charges are placed near each other. Compared to the force on particle X
Art [367]

The force on charge Y is the same as the force on charge X

Explanation:

We can answer this problem by applying Newton's third law of motion, which states that:

"When an object A exerts a force on object B (action force), then object B exerts an equal and opposite force on object A (reaction force)"

In this problem, we can identify object A as charge X and object B as charge Y. The magnitude of the electrostatic force between them is given by

F=k\frac{q_x q_y}{r^2}  (1)

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_x, q_y are the two charges

r is the separation between the two charges

According to Newton's third law, therefore, the magnitude of the force exerted by charge X on charge Y is the  same as the force exerted by charge Y on charge X (and it is given by eq.(1)), however their directions are opposite.

Learn more about Newton's third law:

brainly.com/question/11411375

#LearnwithBrainly

6 0
3 years ago
Any ss2 here (11th Grade)
Sever21 [200]

Answer:

<h2>a) 50°</h2><h2>b) 40°</h2>

Explanation:

Check the complete diagram n the attachment below

a) The angle of incidence on a plane surface is the angle between the incidence ray and the normal ray acting on a plane surface. The normal ray is the ray perpendicular to the surface while the incidence ray is the ray striking a plane surface.

According to the diagram, the angle of reflection r₂ on M₂ is 90°-g where g is the angle of glance.

Given angle of glance on M₂ to be 40°, r₂ = 90-40 = 50°

According the second law of reflection, the angle of incidence = angle of reflection, therefore i₂ = r₂ = 50° (on M₂)

Also ∠OO₂O₁ =  ∠OO₁O₂ = 40° (angle of glance on M₁){alternate angle}

The angle of incidence on M₁ = 90° - 40° = 50°

b) The angle of incidence to the surface of M₁(∠PO₁A)will be the angle of glance on M₁ which is equivalent to 40°

6 0
3 years ago
What relationship exists betwen air resistance and acceleration of falling objects
ololo11 [35]
They both make a thing go faster and slower but the relationship is force.
5 0
4 years ago
Read 2 more answers
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