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1 year ago

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A car initially traveling at 15 m/s North accelerates to 25 m/s North in 4 seconds. The magnitude of the average acceleration is

?

1 answer:

GarryVolchara [31]1 year ago

4 0

The average acceleration of the car as it accelerates from 15 m/s to 25 m/s will be 2.5 m/s².

<h3>What is Average acceleration?</h3>

Average acceleration is the average rate of change of velocity with respect to time. Mathematically -

a = Δv/Δt

Given that a car is traveling at 15 m/s in north direction and in 4 seconds it accelerates to 25 m/s in the north direction only. Therefore, we can write -

initial velocity [u] = 15 m/s

final velocity [v] = 25 m/s

time taken [Δt] = 4 s

The magnitude of the average acceleration can be calculated as follows-

a = Δv/Δt

Δv = v - u = 25 - 15

Δv = 10 m/s

Therefore -

a = 10/4

a = 2.5 m/s².

Therefore, the average acceleration of the car will be 2.5 m/s².

To solve more questions on Acceleration, visit the link below-

brainly.com/question/21300541

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A roller coaster car starts from rest at the top of a hill 15 m high and rolls down to ground level. From there it starts into a

Softa [21]

Answer:

955.5N

Explanation:

The normal force is given by the difference between the centripetal force and gravity at the top of the loop:

$F_N = F_C - F_G = m\frac{v^{2} }{r} - mg$

mass m = 65kg

radius of the loop r = 4m

velocity v = ?

g = 9.8 m/s²

To find the centripetal force, you need to find the velocity of the car at the top of the loop.

Use energy conservation:

$E_{tot}=mgh + \frac{1}{2} mv^{2}$

At the top of the hill:

$E_{tot}= mgh_{hill}$

At the top of the loop:

$E_{tot}=mgh_{loo}_p +\frac{1}{2} m v^{2}$

Setting both energies equal and canceling the mass m gives:

$gh_{hill} = gh_{loo}_p + \frac{1}{2} v^{2}$

Solving for v:

$v^{2} = 2g(h_{hill}-h_{loo}_p)$

Using v in the first equation:

$F_N = \frac{2mg(h_{hill}-h_{loo}_p)}{r} - mg$

$F_N = 955.5N$

7 0

3 years ago

A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s

Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g= $4.67\times 10^{-3} kg$

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1, $m_1=1177g=1.177kg$

Mass of block 2, $m_2=1626 g=1.626 kg$

Velocity of block 1, $v_1=0.681m/s$

(a)

Let velocity of the second block after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

$mv=m_1v_1+(m+m_2)v_2$

$4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2$

$1.66719=0.801537+1.63067v_2$

$1.66719-0.801537=1.63067v_2$

$0.865653=1.63067v_2$

$v_2=\frac{0.865653}{1.63067}$

$v_2=0.531m/s$

Hence, the velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

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

$k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)$

$k_i=297.59 J$

Final kinetic energy after collision

$K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2$

$K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2$

$K_f=0.5028 J$

Now, he ratio of the total kinetic energy after the collision to that before the collision

= $\frac{k_f}{k_i}=\frac{0.5028}{297.59}$

=0.00169

5 0

3 years ago

On which date is the gravitational force between Earth and the moon the greatest?

Genrish500 [490]

The gravitational forces between the Earth and Moon are greatest when the two bodies are closest together. That happens every 27.32 days, when the Moon is at the perigee of its orbit.

Even if this happened at the same time in every orbit, the date would change, because there are not 27.32 days in a month.

But it doesn't happen at the same time in every orbit ... the Moon's perigee precesses around its orbit, on account of the gravitational forces toward the Earth, the Sun, Venus, Mars, and the other planets.

3 0

3 years ago

A passenger train left station A at 6:00 p.m. Moving with the average speed 45 mph, it arrived at station B at 10:00 p.m. A tran

marishachu [46]

<h2>Average speed of transit train is 60 mph</h2>

Explanation:

Average speed of passenger train = 45 mph

Time taken from station A to station B for passenger train = 10:00 - 6:00 = 4 hours

Distance between station A to station B = 45 x 4 = 180 miles.

Time taken from station A to station B for transit train = 4 - 1 = 3 hours

Distance between station A to station B = Average speed of transit train x Time taken from station A to station B for transit train

180 = Average speed of transit train x 3

Average speed of transit train = 60 mph

Average speed of transit train is 60 mph

8 0

4 years ago

The cross section of a copper strip is 1.2 mmthick and 20 mm wide. There is a 25-A current through this cross section, with the

Naily [24]

To solve this problem it is necessary to use the concepts related to the Hall Effect and Drift velocity, that is, at the speed that an electron reaches due to a magnetic field.

The drift velocity is given by the equation:

$V_d = \frac{I}{nAq}$

Where

I = current

n = Number of free electrons

A = Cross-Section Area

q = charge of proton

Our values are given by,

$I = 25 A$

$A= 1.2*20 *10^{-6} m^2$

$q= 1.6*10^{-19}C$

$N = 8.47*10^{19} mm^{-3}$

$V_d =\frac{25}{(1.2*20 *10^{-6})(1.6*10^{-19})(8.47*10^{19} )}$

$V_d = 7.68*10^{-5}m/s$

The hall voltage is given by

$V=\frac{IB}{ned}$

Where

B= Magnetic field

n = number of free electrons

d = distance

e = charge of electron

Then using the formula and replacing,

$V=\frac{(2.5)(25)}{(8.47*10^{28})(1.6*10^{-19})(1.2*10^{-3})}$

$V = 3.84*10^{-6}V$

5 0

3 years ago