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Lubov Fominskaja [6]
3 years ago
15

A car that experiences no frictional force is started and caused to move. For the car to continue in that motion, the gas pedal

would have to be used:
Physics
1 answer:
astraxan [27]3 years ago
8 0

Explanation:

Gas pedal is not required to used on friction less surface. On a friction less surface once the car is started and caused to move, car continues to move with same velocity as there is no opposing force. Therefore, no pressing of gas pedal required. However, this situation is not possible in real life, as friction is always present.

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A flatbed truck is carrying a 20.0 kg crate along a level road. the coefficient of static friction between the crate and the bed
Dahasolnce [82]
<span>3.92 m/s^2 Assuming that the local gravitational acceleration is 9.8 m/s^2, then the maximum acceleration that the truck can have is the coefficient of static friction multiplied by the local gravitational acceleration, so 0.4 * 9.8 m/s^2 = 3.92 m/s^2 If you want the more complicated answer, the normal force that the crate exerts is it's mass times the local gravitational acceleration, so 20.0 kg * 9.8 m/s^2 = 196 kg*m/s^2 = 196 N Multiply by the coefficient of static friction, giving 196 N * 0.4 = 78.4 N So we need to apply 78.4 N of force to start the crate moving. Let's divide by the crate's mass 78.4 N / 20.0 kg = 78.4 kg*m/s^2 / 20.0 kg = 3.92 m/s^2 And you get the same result.</span>
6 0
3 years ago
What is the speed of a transverse wave in a rope of length 3. 1 m and mass 86 g under a tension of 380 n?
Nitella [24]

117 m/sec is the speed of a transverse wave in a rope of length 3. 1 m and mass 86 g under a tension of 380 n.

The wave speed v is given by

v= √τ/μ

​where τ is the tension in the rope and μ is the linear mass density of the rope.

The linear mass density is the mass per unit length of rope :

μ= m / L = (0.086 kg)/(3.1 m)=0.0277 kg/m.

v= \sqrt{ \frac{380 N}{0.0277 kg/m}}  = 117.125 m/sec (approx. 117 m/sec

In physics, a transverse wave is a wave whose oscillations are perpendicular to the direction of the wave's advance. This is in contrast to a longitudinal wave which travels in the direction of its oscillations. Water waves are an example of transverse wave.

Transverse waves commonly occur in elastic solids due to the shear stress generated; the oscillations in this case are the displacement of the solid particles away from their relaxed position, in directions perpendicular to the propagation of the wave. These displacements correspond to a local shear deformation of the material. Hence a transverse wave of this nature is called a shear wave. Since fluids cannot resist shear forces while at rest, propagation of transverse waves inside the bulk of fluids is not possible.

Learn more about Transverse waves here : brainly.com/question/13761336

#SPJ4

3 0
2 years ago
In the picture of a planet in orbit around its star area a is double that of area b. What is true of the time the planet takes t
aalyn [17]
The Kepler's laws predict the planetary motion, so there are three laws for this, namely:

1. The orbit of a planet is an ellipse with the Sun (the sun is a star!) at one of the two focus.

2. A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.

3. The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit.

So, let's use second law. The Sun sweeps out equal areas during equal intervals of time means that if A = B, the time the planet takes to travel A1A2 is equal to the time the planet takes to travel B1B2, but given that A = 2B, then takes twice the time to travel A1A2 compared to B1B2.
8 0
3 years ago
At t=0 bullet A is fired vertically with an initial (muzzle) velocity of 450 m/s. When 3s. bullet B is fired upward with a muzzl
Debora [2.8K]

Answer:

At time 10.28 s after A is fired bullet B passes A.

Passing of B occurs at 4108.31 height.

Explanation:

Let h be the height at which this occurs and t be the time after second bullet fires.

Distance traveled by first bullet can be calculated using equation of motion

s=ut+0.5at^2 \\

Here s = h,u = 450m/s a = -g and t = t+3

Substituting

h=450(t+3)-0.5\times 9.81\times (t+3)^2=450t+1350-4.9t^2-29.4t-44.1\\\\h=420.6t-4.9t^2+1305.9

Distance traveled by second bullet

Here s = h,u = 600m/s a = -g and t = t

Substituting

h=600t-0.5\times 9.81\times t^2=600t-4.9t^2\\\\h=600t-4.9t^2 \\

Solving both equations

600t-4.9t^2=420.6t-4.9t^2+1305.9\\\\179.4t=1305.9\\\\t=7.28s \\

So at time 10.28 s after A is fired bullet B passes A.

Height at t = 7.28 s

h=600\times 7.28-4.9\times 7.28^2\\\\h=4108.31m \\

Passing of B occurs at 4108.31 height.

6 0
3 years ago
Who’s of the following is classified as a salt: ammonia or sodium chloride?
Temka [501]

Sodium Chloride could be classified as a salt.

8 0
3 years ago
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