All categories

2 years ago

A car accelerates at a constant rate from 0 to 50 mph in three fourths min. How far does the car travel during that time?

Physics

1 answer:

Helen [10]2 years ago

8 0

Answer:

the car have travelled 0.31 mile during that time

Explanation:

Applying the Equation of motion;

s = 0.5(u+v)t

Where;

s = distance travelled

u = initial speed = 0 mph

v = Final speed = 50 mph

t = time taken = 3/4 min = 3/4 ÷ 60 hours = 1/80 hour

Substituting the given values into the equation;

s = 0.5(0+50)×(1/80)

s = 0.3125 miles

s ~= 0.31 mile

the car have travelled 0.31 mile during that time

You might be interested in

A bullet is fired horizontally at a height of 1.3 meters and a velocity of 950 m/s. How long was the bullet in the air?

seropon [69]

Answer:

<em>The bullet was 0.52 seconds in the air.</em>

Explanation:

<u>Horizontal Motion </u>

It occurs when an object is thrown horizontally with a speed v from a height h.

The object describes a curved path ruled exclusively by gravity until it hits the ground.

To calculate the time the object takes to hit the ground, we use the following equation:

$\displaystyle t=\sqrt{\frac{2y}{g}}$

Note it doesn't depend on the initial velocity but on the height.

The bullet is fired horizontally at h=1.3 m, thus:

$\displaystyle t=\sqrt{\frac{2\cdot 1.3}{9.8}}$

$\displaystyle t=\sqrt{\frac{2.6}{9.8}}$

t = 0.52 s

The bullet was 0.52 seconds in the air.

3 0

2 years ago

An object is accelerating from rest at a rate of 3 m/s2.How long will it take the object to attain 18 m/s ? *

Temka [501]

Initial velocity =0, a=3m/s2, final velocity =18m/s, 18=3t, t=6 sec

7 0

2 years ago

Write this large number in scientific notation. Determine the values of Vm) and (n) when the following mass of the Earth is writ

hichkok12 [17]

Answer : The answer is, 5.97, 24

Explanation :

Scientific notation : It is the representation of expressing the numbers that are too big or too small and are represented in the decimal form with one digit before the decimal point times 10 raise to the power.

For example :

5000 is written as $5.0\times 10^3$

889.9 is written as $8.899\times 10^{-2}$

In this examples, 5000 and 889.9 are written in the standard notation and $5.0\times 10^3$ and $8.899\times 10^{-2}$ are written in the scientific notation.

If the decimal is shifting to right side, the power of 10 is negative and if the decimal is shifting to left side, the power of 10 is positive.

As we are given the 5,970,000,000,000,000,000,000,000 in standard notation.

Now converting this into scientific notation, we get:

$\Rightarrow 5,970,000,000,000,000,000,000,000=5.97\times 10^{24}$

As, the decimal point is shifting to left side, thus the power of 10 is positive.

Hence, the answer is, $5.97\times 10^{24}$

Now the answer is comparing to $m.\times 10^n$

So, m = 5.97 and n = 24

Thus, the answer is, 5.97, 24

5 0

2 years ago

(b) The output current of the generator is 2.0A. The output potential difference (p.d.) of the

Luden [163]

Answer:

P = 22 watts

Explanation:

Given that,

The output power of the generator = 2 a

The output potential difference = 11 V

We need to find the output power of the generator. The formula for the output power is given by :

$P=V\times I\\\\P=11\times 2\\\\P=22\ W$

So, the output power of the generator is equal to 22 Watts.

8 0

2 years ago

two horses are pulling a 325 kg wagon, initially at rest. The horses exert 250 N and 178 N forward forces, respectively. Ignorin

Dovator [93]

Answer:

AFter 3.5 s, the wagon is moving at: $4.62\,\,\frac{m}{s}$

Explanation:

Let's start by finding first the net force on the wagon, and from there the wagon's acceleration (using Newton's 2nd Law):

Net force = 250 N + 178 N = 428 N

Therefore, the acceleration from Newton's 2nd Law is:

$F=m\,*\,a\\a = \frac{F}{m} \\a= \frac{428}{325}\, \frac{m}{s^2} \\a\approx 1.32 \,\,\frac{m}{s^2}$

So now we apply this acceleration to the kinematic expression for velocity in an object moving under constant acceleration:

$v_f=v_i+a\,*\,t\\v_f=0+1.32\,*\,3.5\,\,\frac{m}{s} \\v_f=4.62\,\,\frac{m}{s}$

7 0

3 years ago

A car accelerates at a constant rate from 0 to 50 mph in three fourths min. How far does the car travel during that​ time?

A car accelerates at a constant rate from 0 to 50 mph in three fourths min. How far does the car travel during that time?