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2 years ago

What is the net force on a car moving in a straight line with a constant velocity

Physics

2 answers:

guapka [62]2 years ago

3 0

Answer:

For example, when a car travels at a constant speed, the driving force from the engine is balanced by resistive forces such as air resistance and friction in the car's moving parts. The resultant force on the car is zero.

Explanation:

hope this helps

vfiekz [6]2 years ago

3 0

Answer:

zero

Explanation:

The acceleration of the car is zero, and and in this case the velocity is also zero. When your car is moving at constant velocity down a street, the net force must also be zero, according to Newton's First Law.

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Velocityis a vector quantity which has both magnitude and direction. Using complete sentences, describethe object's velocity. Co

mr_godi [17]

Answer:

Explanation:

A physical quantity which can be completely described by the magnitude and direction both are called vector quantities. For example, displacement, velocity, etc.

A physical quantity which can be completely explained by the magnitude only is called scalar quantity. For example, mass, time, etc.

8 0

3 years ago

A helicopter starting on the ground is rising directly into the air at a rate of 25 ft/s. You are running on the ground starting

rusak2 [61]

Answer:

The rate of change of the distance between the helicopter and yourself (in ft/s) after 5 s is $\sqrt{725}$ ft/ sec

Explanation:

Given:

h(t) = 25 ft/sec

x(t) = 10 ft/ sec

h(5) = 25 ft/sec . 5 = 125 ft

x(5) = 10 ft/sec . 5 = 50 ft

Now we can calculate the distance between the person and the helicopter by using the Pythagorean theorem

$D(t) = \sqrt{h^2 + x^2}$

Lets find the derivative of distance with respect to time

$\frac{dD}{dt} (t) = \frac{2h \cdot \frac{dh}{dt} +2x \cdot\frac{dx}{dt}} {2\sqrt{h^2 + x^2}}$

Substituting the values of h(t) and x(t) and simplifying we get,

$\frac{dD}{dt}(t) = \frac{50t \cdot \frac{dh}{dt} + 20 \cdot \frac{dx}dt}{2\sqrt{625\cdot t^2 + 100 \cdot t^2}}$

$\frac{dh}{dt} = 25ft/sec$

$\frac{dx}{dt} = 10 ft/sec$

$\frac{Dd}{dt} (t) = \frac{1250t +200t}{2\sqrt{725}t}$ = $\frac{725}{\sqrt{725}}$ = $\sqrt{725}$ ft / sec

5 0

3 years ago

A small airplane has to reach a speed if 27.8 m/s to takeoff. It can accelerate at 2.00 m/s^2. What is the minimal length of run

pickupchik [31]

Using the constant acceleration formula v^2 = u^2 + 2as, we can figure out that it would take a distance of 193.21m to reach 27.8m/s

3 0

3 years ago

In order for an object to have kinetic energy it must have a mass and a ?

almond37 [142]

Answer:

Velocity

Explanation:

The mechanical energy of the body is defined as the sum of the potential energy and kinetic energy.

E = P.E + K.E

The potential energy of a body is due to the height from the surface of the earth.

P.E = mgh

The kinetic energy of the is possessed by the body due to the virtue of its motion,

K.E = ½ mv²

If there is no velocity associated with the body, there is no K.E in the body.

8 0

3 years ago

If a 1000-pound capsule weighs only 165 pounds on the moon, how much work is done in propelling this capsule out of the moon's g

Bas_tet [7]

Answer:

W = 1,307 10⁶ J

Explanation:

Work is the product of force by distance, in this case it is the force of gravitational attraction between the moon (M) and the capsule (m₁)

F = G m₁ M / r²

W = ∫ F. dr

W = G m₁ M ∫ dr / r²

we integrate

W = G m₁ M (-1 / r)

We evaluate between the limits, lower r = R_ Moon and r = ∞

W = -G m₁ M (1 /∞ - 1 / R_moon)

W = G m1 M / r_moon

Body weight is

W = mg

m = W / g

The mass is constant, so we can find it with the initial data

For the capsule

m = 1000/32 = 165 / g_moon

g_moom = 165 32/1000

.g_moon = 5.28 ft / s²

I think it is easier to follow the exercise in SI system

W_capsule = 1000 pound (1 kg / 2.20 pounds)

W_capsule = 454 N

W = m_capsule g

m_capsule = W / g

m = 454 /9.8

m_capsule = 46,327 kg

Let's calculate

W = 6.67 10⁻¹¹ 46,327 7.36 10²² / 1.74 10⁶

W = 1,307 10⁶ J

7 0

3 years ago