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3 years ago

A car is traveling at 10 m/s. 10 seconds later the car is traveling 40 m/s. What is the car’s acceleration?

Physics

1 answer:

ratelena [41]3 years ago

3 0

Answer:

a = 3 m/s^2

Explanation:

Vi = 10 m/s

Vf = 40 m/s

t = 10 s

Plug those values into the following equation:

Vf = Vi + at

40 = 10 + 10a

---> a = 3 m/s^2

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Suppose the electric field in problems 2 was caused by a point charge. The test charge is moved to a distance twice as far from

mojhsa [17]

Answer:

it is reduced four times.

Explanation:

By definition, the electric field is the force per unit charge created by a charge distribution.

If the charge creating the field is a point charge, the force exerted by it on a test charge, must obey Coulomb´s Law, so, it must be inversely proportional to the square of the distance between the charges.

So, if the distance increases twice, as the force is inversely proportional to the square of the distance, and the square of 2 is 4, this means that the magnitude of the force exerted on the test charge must be 4 times smaller.

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3 years ago

A bomb of mass 6kg initially at rest explodes into two fragments of masses 4kg and2kg respectively. If the greater mass moves wi

Katen [24]

Answer:

v = 10 [m/s]

Explanation:

The largest mass is that of 4 [kg], in this way the momentum can be calculated by means of the product of the mass by velocity.

$P=m*v\\$

where:

P = momentum [kg*m/s]

m = mass = 4 [kg]

v = velocity = 5 [m/s]

Now the momentum:

$P=4*5\\P=20[kg*m/s]$

This same momentum is equal for the other mass, in this way we can find the velocity.

$P=m*v\\20=2*v\\v=10[m/s]$

7 0

3 years ago

A 1200-kg car initially at rest undergoes constant acceleration for 9.4 s, reaching a speed of 11 m/s. It then collides with a s

Natalka [10]

To solve this problem we will apply the principle of conservation of energy and the definition of kinematic energy as half the product between mass and squared velocity. So,

$KE_i = KE_f$

$KE_f = \frac{1}{2} mv^2$

Here,

m = Mass

V = Velocity

Replacing,

$KE_f = \frac{1}{2} (12000)(11)^2$

$KE_f = 72600J$

Therefore the final kinetic energy of the two car system is 72.6kJ

8 0

3 years ago

A 60kg bicyclist (including the bicycle) is pedaling to the

Fittoniya [83]

a) 4 forces

b) 186 N

c) 246 N

Explanation:

Let's count the forces acting on the bicylist:

1) Weight ( $W=mg$ ): this is the gravitational force exerted on the bicyclist by the Earth, which pulls the bicyclist towards the Earth's centre; so, this force acts downward (m = mass of the bicyclist, g = acceleration due to gravity)

2) Normal reaction (N): this is the reaction force exerted by the road on the bicyclist. This force acts vertically upward, and it balances the weight, so its magnitude is equal to the weight of the bicyclist, and its direction is opposite

3) Applied force ( $F_A$ ): this is the force exerted by the bicylicist to push the bike forward. Its direction is forward

4) Air drag ( $R$ ): this is the force exerted by the air on the bicyclist and resisting the motion of the bike; its direction is opposite to the motion of the bike, so it is in the backward direction

So, we have 4 forces in total.

Here we can find the net force on the bicyclist by using Newton's second law of motion, which states that the net force acting on a body is equal to the product between the mass of the body and its acceleration:

$F_{net}=ma$

where

$F_{net}$ is the net force

m is the mass of the body

a is its acceleration

In this problem we have:

m = 60 kg is the mass of the bicyclist

$a=3.1 m/s^2$ is its acceleration

Substituting, we find the net force on the bicyclist:

$F_{net}=(60)(3.1)=186 N$

We can write the net force acting on the bicyclist in the horizontal direction as the resultant of the two forces acting along this direction, so:

$F_{net}=F_a-R$

where:

$F_{net}$ is the net force

$F_a$ is the applied force (forward)

$R$ is the air drag (backward)

In this problem we have:

$F_{net}=186 N$ is the net force (found in part b)

$R=60 N$ is the magnitude of the air drag

Solving for $F_a$ , we find the force produced by the bicyclist while pedaling:

$F_a=F_{net}+R=186+60=246 N$

3 0

3 years ago

Brainliest for correct answers :)

Nostrana [21]

Answer:

a. 1100 meters.

b. Between B and C

c.1. Between point D and E

c2. Between point D and E

d. 3.7 m/s.

Explanation:

The girl travels the distance of 1100 meters from starting to the end. There is no motion occurs between B and C due to no change of distance value from 200 meters. Between point D and E, the girls covers 500 meters long distance and also covers fastest distance between point D and E because between point D and E, the girl covers 500 meters distance in 30 seconds which is the highest of all. The average speed of the girls is 3.7 meter/seconds if we divide total distance i.e. 1100 meters by time which is 300 seconds.

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3 years ago