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True [87]
3 years ago
6

A force of 3,000 N is applied to a car, resulting in acceleration. In 10 s, the car increases its velocity from 0 m/s to 30 m/s.

What is the mass of the car?​
Physics
1 answer:
Reil [10]3 years ago
6 0

Answer:

4.5 m/s

Explanation:

I am not sure if it correct.

You might be interested in
A boat moves with a speed of +2.5 m/s in a direction 25° north of east. If the mass of the boat is 15,000 kg, what is the moment
nadya68 [22]
Momentum - mass in motion
P=MV
P=(15,000 kg)(2.5 m/s)
P=37 500 kg x m/s to the north
Hope this helps
5 0
3 years ago
Estimate how far apart the rays of deepest red and deepest violet light are as they exit the bottom surface. assume nred = 1.57
Harlamova29_29 [7]
We begin by noting that the angle of incidence is the one that's taken with respect to the normal to the surface in question. In this case the angle of incidence is 30. The material is Flint Glass according to the original question. The refractive indez of air n1=1, the refractive index of red in flint glass is nred=1.57, finally for violet in the glass medium is nviolet=1.60. Snell's Law dictates:
n_1sin(\theta_1)=n_2sin(\theta_2)
Where \theta_2 differs for each wavelenght, that means violet and red will have different refractive indices in the glass.
In the second figure provided details are given on which are the angles in question, \Delta x is the distance between both rays.
\theta_{2red}=Asin(\frac{sin(30)}{1.57})\approx 18.5705
\theta_{2violet}=Asin(\frac{sin(30)}{1.60})\approx 18.21
At what distance d from the incidence normal will the beams land at the bottom?
For violet we have:
d_{violet}=h.tan(\theta_{2violet})\approx 0.0132m
For red we have:
d_{red}=h.tan(\theta_{2red})\approx 0.0134m
We finally have:
\Delta x=d_{red}-d_{violet}\approx2.8\times10^{-4}m


6 0
3 years ago
You and a friend each hold a lump of wet clay. Each lump has a mass of 30 grams. You each toss your lump of clay into the air, w
Vesna [10]

Answer:

\ \text{m/s}

Explanation:

u_1 = Velocity of one lump = 3x+3y-3z

u_2 = Velocity of the other lump = -4x+0y-4z

m = Mass of each lump = 30\ \text{g}

The collision is perfectly inelastic as the lumps stick to each other so we have the relation

mu_1+mu_2=(m+m)v\\\Rightarrow m(u_1+u_2)=2mv\\\Rightarrow v=\dfrac{u_1+u_2}{2}\\\Rightarrow v=\dfrac{3x+3y-3z-4x+0y-4z}{2}\\\Rightarrow v=-0.5x+1.5y-3.5z=\ \text{m/s}

The velocity of the stuck-together lump just after the collision is \ \text{m/s}.

4 0
2 years ago
Man made objects that spread an impulse over a large amount of time
lyudmila [28]

Answer:

Examples of man-made objects that spread an impulse over a large amount of time are "airbags" in vehicles and "arrestor beds" (for trucks).

Explanation:

The question above is highly related to the topic about "Impulse" in Physics.

"Impulse"<em> refers to an object's change in momentum (the amount of motion in an object) when a force acts upon it for an interval time.</em> When it comes to providing safety to people when it comes to vehicular crashes, impulse plays a vital role.

Let's take the example of airbags in vehicles. Once a vehicle collides with another object, the driver is carried by a forward motion. Without airbags, the time is normally shorter for the driver to be stopped by the windshield. This results to a greater force. However, with the presence of air-bags, the driver will hit the airbag, instead of the windshield. <u>This will lengthen the time of the impact, thus reducing the force.</u>

Another example are the arrestor beds for trucks. Arrestor beds have been designed in order for trucks to stop, since it's hard to maneuver them. <u>With the help of arrestor beds, trucks are able to come to a stop with a longer time interval, but decreased force.</u>

5 0
3 years ago
What is the torque τa about axis a due to the force f⃗ ? express the torque about axis a at cartesian coordinates (0,0)?
pentagon [3]

Answer:

\tau_a = F a sin \theta

Explanation:

The torque of a force is given by:

\tau = F d sin \theta

where

F is the magnitude of the force

d is the distance between the point of application of the force and the centre of rotation of the system

\theta is the angle between the direction of the force and d

In this problem, we have:

F, the force

a, the distance of application of the force from the centre (0,0)

\theta, the angle between the direction of the force and a

Therefore, the torque is

\tau_a = F a sin \theta

5 0
3 years ago
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