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yKpoI14uk [10]
3 years ago
13

A car is traveling at vx = 30 m/s . The driver applies the brakes and the car decelerates at ax = -5.0 m/s2 . What is the stoppi

ng distance?
Physics
1 answer:
____ [38]3 years ago
6 0

Answer:

90 m

Explanation:

We use an ecuation of uniformly accelerated motion, which allows us to find the distance traveled by the car from the moment that driver applies the brakes until it stops completely, that is, when its final speed is zero.

v_{f}^2=v_{o}^2+2ad

We isolate the variable d, knowing that the final speed v_{f} is zero

d=\frac{-v_{o}^2}{2a}\\d=\frac{-(30m/s)^2}{2(-5m/s^2)}=90 m

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A point charge of 6.8 C moves at 6.5 × 104 m/s at an angle of 15° to a magnetic field that has a field strength of 1.4 T.
Gekata [30.6K]

As per the question, the point charge is given as [q] = 6.8 C

The velocity of the charged particle [v] = 6.5*10^{4}\ m/s

The magnetic field [B] = 1.4 T

The angle made between magnetic field and velocity [\theta] = 15 degree.

We are asked to calculate the magnetic force experienced by the charged  particle.

The magnetic force experienced by the charged particle is calculated as -

Magnetic force \vec F=q(\ \vec V \times \vec B)

                        i.e F = =qVBsin\theta

                                   =6.8*6.5*10^{4}*1.4*sin15

                                   =61.88*10^{4}sin15

                                   =61.88*10^4*0.2588\ N

                                   =16.016*10^4\ N

                                   =1.6*10^5\ N

Hence, the force experienced by the charged particle is C i.e 1.6*10^5 N

                                   

             

4 0
3 years ago
Read 2 more answers
A 1000-kg car is slowly picking up speed as it goes around a horizontal curve whose radius is 100 m. The coefficient of static f
Snezhnost [94]

Answer:

18.5 m/s

Explanation:

On a horizontal curve, the frictional force provides the centripetal force that keeps the car in circular motion:

\mu mg = m\frac{v^2}{r}

where

\mu is the coefficient of static friction between the tires and the road

m is the mass of the car

g is the gravitational acceleration

v is the speed of the car

r is the radius of the curve

Re-arranging the equation,

v=\sqrt{\mu gr}

And by substituting the data of the problem, we find the speed at which the car begins to skid:

v=\sqrt{(0.350)(9.8 m/s^2)(100 m)}=18.5 m/s

7 0
3 years ago
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What is the potential energy for and object with mass of 15 Kg and 20 m above the ground?
vaieri [72.5K]

Answer:

3000J

Explanation:

The gravitational energy for any object is given by E=mgh

where m is mass, g is gravitational field strength anf h is height above ground

5 0
3 years ago
This equation goes with which law?
Alchen [17]

Answer:

I think that's Newton's second law of motion

Explanation:

f = m(v-u)

________

t

since a = (v-u)t

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3 0
3 years ago
If you were standing at the center of curvature in front of a concave mirror, what image would be projected?
Romashka [77]
When you stand at the center of curvature in front of a concave mirror, the image created will be formed at the center of curvature also, but it will be inverted. The image formed will be real, and will have the same size as you.
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