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3 years ago

If the brakes on a 1,600 kg car exert a stopping force of

Physics

1 answer:

ra1l [238]3 years ago

8 0

The time it takes the car to stop is 1.28 seconds.

<h3 /><h3>What is impulse</h3>

Impulse can be defined as the product of force and time.

To calculate the time it will take the car to stop. we apply the formula of impulse.

Formula:

Ft = m(v-u)............ Equation 1

Where:

F = Force on the car
m = mass of the car
v = final velocity
u = initial velocity.
t = time

Make t the subject of the equation

t = m(v-u)/F.............. Equation 2

From the question,

Given:

F = -25000 N
v = 0 m/s
u = 20 m/s
m = 1600 kg

Substitute these values into equation 2

t = 1600(0-20)/-25000
t = -3200/-2500
t = 1.28 seconds.

Hence, the time it takes the car to stop is 1.28 seconds.

Learn more about impulse of a force here: brainly.com/question/20586658

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Scicene because it has to have a body scored so yea science

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3 years ago

An object on the surface of the earth weighs 90 lb. At two earth radii above the surface, It will welgh:

vredina [299]

I have no explanation, but saw this pop up on a test paper and the answe was 10lb

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3 years ago

The wavelength of a wave on a string is 1.2 meters. If the speed of the wave is 60 meters/second, what is its frequency? A. 0.20

anastassius [24]

Answer:

50 Hz

Explanation:

The frequency of a wave is given by

$f=\frac{v}{\lambda}$

where

v is the speed of the wave

$\lambda$ is the wavelength

In this problem,

v = 60 m/s

$\lambda=1.2 m$

So the frequency is

$f=\frac{60 m/s}{1.2 m}=50 Hz$

8 0

3 years ago

a bus is moving with the velociity of 36 km/hr . after seeing a boy at 20 m ahead on the road, the driver applies the brake and

Klio2033 [76]

Answer:

Assumption: the acceleration of this bus is constant while the brake was applied.

Acceleration of this bus: approximately $\left(-6.0\; \rm m \cdot s^{-2}\right)$ .

It took the bus approximately $1.7\;\rm s$ to come to a stop.

Explanation:

Quantities:

Displacement of the bus: $x = 10\; \rm m$ .
Initial velocity of the bus: $\displaystyle u = 36\; \rm km \cdot hr^{-1} = 36\; \rm km \cdot hr^{-1}\times \frac{1\; \rm m \cdot s^{-1}}{3.6\; \rm km\cdot hr^{-1}} = 10\; \rm m \cdot s^{-1}$ .
Final velocity of the bus: $v = 0\; \rm m\cdot s^{-1}$ because the bus has come to a stop.
Acceleration, $a$ : unknown, but assumed to be a constant.
Time taken, $t$ : unknown.

Consider the following SUVAT equation:

$\displaystyle x = \frac{1}{2}\, \left(a\, t^2\right) + u\, t$ .

On the other hand, assume that the acceleration of this bus is indeed constant. Given the initial and final velocity, the time it took for the bus to stop would be inversely proportional to the acceleration of this bus. That is:

$\displaystyle t = \frac{v - u}{a}$ .

Therefore, replace the quantity $t$ with the expression $\displaystyle \left(\frac{v - u}{a}\right)$ in that SUVAT equation:

$\displaystyle x = \frac{1}{2}\, \left(a\, \left(\frac{v -u}{a}\right)^2\right) + u\, \left(\frac{v - u}{a}\right)$ .

Simplify this equation:

$\begin{aligned}x &= \frac{1}{2}\, \left(a\, {\left(\frac{v -u}{a}\right)}^2\right) + u\, \left(\frac{v - u}{a}\right) \\ &= \frac{1}{2}\left(\frac{{(v - u)}^2}{a}\right) + \frac{u\, (v - u)}{a} =\frac{1}{a}\, \left(\frac{{(v - u)}^2}{2} + u\, (v - u)\right)\end{aligned}$ .

Therefore, $\displaystyle a= \frac{1}{x}\, \left(\frac{{(v - u)}^2}{2} + u\, (v - u)\right)$ .

In this question, the value of $x$ , $u$ , and $v$ are already known:

$x = 10\; \rm m$ .
$\displaystyle u =10\; \rm m \cdot s^{-1}$ .
$v = 0\; \rm m\cdot s^{-1}$ .

Substitute these quantities into this equation to find the value of $a$ :

$\begin{aligned} a &= \frac{1}{x}\, \left(\frac{{(v - u)}^2}{2} + u\, (v - u)\right) \\ &= \frac{1}{10\; \rm m}\times \left(\frac{{\left(0\; \rm m \cdot s^{-1} - 10\; \rm m \cdot s^{-1}\right)}^2}{2} + \left(0\; \rm m \cdot s^{-1} - 10\; \rm m \cdot s^{-1}\right)\times 10\; \rm m \cdot s^{-1}\right)\\ &\approx -6.0\; \rm m \cdot s^{-2}\end{aligned}$ .

(The value of acceleration $a$ is less than zero because the velocity of the bus was getting smaller.)

Substitute $a \approx -6.0\; \rm m \cdot s^{-2}$ (alongside $u = 10\; \rm m \cdot s^{-1}$ and $v = 0\; \rm m \cdot s^{-1}$ ) to estimate the time required for the bus to come to a stop:

$\begin{aligned}t &= \frac{v - u}{a} \\ &\approx \frac{0\; \rm m \cdot s^{-1} - 10\; \rm m \cdot s^{-1}}{-6.0\; \rm m \cdot s^{-2}} \approx 1.7\; \rm s\end{aligned}$ .

8 0

3 years ago

imagine that a tank is filled with water the hight of the liquid column is 7 meters and the area is 1.5 sq meters (m™). what's t

Annette [7]

Answer - 12,900 Newtons

Explanation

First, we find the volume of the water
Volume = Area * Heinght
= 1.5 m² x 7 m
= 10.5 m³

Covert the volume to liters
1 m³ of water = 1000 liters
10.5 m³ of water = 10.5 m³ * 1000 liters liter/m³
= 10,500 liters

Use the volume of water to calculate the mass
1 liter of water weighs 1 kg
10,500 liters of water = 10,500 * 1 kg/liter
= 10,500 kg

Now, we can calculate the force of gravity on the water
Force of gravity on the water = Weight of the water
Weight = Mass * Acceleration

Mass = 10,500kg
Acceleration (due to gravity) = 9.8 m/s²

Force of gravity on the water
= Weight of the water
= Mass * Acceleration
= 10,500 kg * 9.8 m/s²
= 102,900 Newtons

8 0

3 years ago