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2 years ago

A car going 30 m/s to 70 m/s in 10 seconds calculate the acceleration of the car

Physics

2 answers:

Zina [86]2 years ago

7 0

70m/s-30m/s is eqal to 40m/s. Divide 40 by 10 and you get 4. This would be the accleration of the car.

Anastasy [175]2 years ago

6 0

4 should be the answer for the acceleration for the car. I hope this helps!

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In a 400-m relay race the anchorman (the person who runs the last 100 m) for team A can run 100 m in 9.8 s. His rival, the ancho

melisa1 [442]

Answer:

largest lead = 3 m

Explanation:

Basically, this problem is about what is the largest possible distance anchorman for team B can have over the anchorman for team A when the final leg started that anchorman for team A won the race. This show that anchorman for team A must have higher velocity than anchorman for team B to won the race as at the starting of final leg team B runner leads the team A runner.

So, first we need to calculate the velocities of both the anchorman

given data:

Distance = d = 100 m

Time arrival for A = 9.8 s

Time arrival for B = 10.1 s

Velocity of anchorman A = D / Time arrival for A

=100/ 9.8 = 10.2 m/s

Velocity of anchorman B = D / Time arrival for B

=100/10.1 = 9.9 m/s

As speed of anchorman A is greater than anchorman B. So, anchorman A complete the race first than anchorman B. So, anchorman B covered lower distance than anchorman A. So to calculate the covered distance during time 9.8 s for B runner, we use

d = vt

= 9.9 x 9.8 = 97 m

So, during the same time interval, anchorman A covered 100 m distance which is greater than anchorman B distance which is 97 m.

largest lead = 100 - 97 = 3 m

So if his lead no more than 3 m anchorman A win the race.

5 0

2 years ago

You see a total eclipse of the sun. Three Saros cycles later, you find yourself at the same location.

alukav5142 [94]

Answer:

The correct answer is B. You would not have to travel far to see another total eclipse .

Explanation:

it is said that the way to predict or know beforehand if there is a lunar or solar eclipse is just to know the what is called the saros which are periods of exactly 223 synodic months and the eclipses are repeated or occurred in the same location every 3 Saros cycles.

5 0

2 years ago

A 500 kg dragster accelerates from rest to a final speed of 100 m/s in 400 m (about a quarter of a mile) and encounters an avera

Fantom [35]

In order to accelerate the dragster at a speed $v_f = 100 m/s$ , its engine must do a work equal to the increase in kinetic energy of the dragster. Since it starts from rest, the initial kinetic energy is zero, so the work done by the engine to accelerate the dragster to 100 m/s is
$W= K_f - K_i = K_f = \frac{1}{2}mv_f^2=2.5 \cdot 10^6 J$

however, we must take into account also the fact that there is a frictional force doing work against the dragster, and the work done by the frictional force is:
$W_f = F_f d = -(1200 N)(400 m)= -4.8 \cdot 10^5 J$
and the sign is negative because the frictional force acts against the direction of motion of the dragster.

This means that the total work done by the dragster engine is equal to the work done to accelerate the dragster plus the energy lost because of the frictional force, which is $-W_f$ :
$W_t = W + (-W_f)=2.5 \cdot 10^6 J+4.8 \cdot 10^5 J=2.98 \cdot 10^6 J$

So, the power delivered by the engine is the total work divided by the time, t=7.30 s:
$P= \frac{W}{t}= \frac{2.98 \cdot 10^6 J}{7.30 s}=4.08 \cdot 10^6 W$

And since 1 horsepower is equal to 746 W, we can rewrite the power as
$P=4.08 \cdot 10^6 W \cdot \frac{1 hp}{746 W} =547 hp$

3 0

3 years ago

Read 2 more answers

A mass of 0.450 kg rotates at constant speed with a period of 1.45 s at a radius R of 0.140 m in the apparatus used in this labo

Mamont248 [21]

Answer:

1.603 s

Explanation:

Given that

Initial mass, = 0.45 kg

Initial period, = 1.45 s

Initial radius, = 0.14 m

Final mass, = 0.55 kg

Final period, = ?

Final radios, = 0.14 m

Since we are finding the rotation period of two masses of same radius, we can assume that the outward force is the same in both cases. This means that

m₁r₁ω₁² = m₂r₂ω2²

Where, ω = 2π/T, on substituting, we have

0.45 * 0.14 * (2π / 1.45)² = 0.550 * 0.14 * (2π / T₂)²

0.45 / 1.45² = 0.550 / T₂²

T₂² = 0.550 * 1.45² / 0.45

T₂² = 2.56972

T₂ = √2.56972

T₂ = 1.603 sec

7 0

2 years ago

Discuss 2 ways tolerance and empathy counteract discrimination in school and community<br>

VladimirAG [237]

Answer:

Reducing Discrimination by Changing Social Norms

Reducing Prejudice through Intergroup Contact

Explanation:

3 0

2 years ago