All categories

3 years ago

What average force is required to stop a 980 kg car in 9.0 s if the car is traveling at 91 km/h ?

1 answer:

5 0

Newton's second law of motion:
F=ma
$a = \frac{v}{t}$
You convert from km/h to m/s by dividing by 3.6:
$v = 91 \div 3.6$
$v = 25.3 \: \frac{m}{s}$
Then a is:
$a = \frac{25.3}{9}$
$a = 2.8 \frac{m}{s {}^{2} }$
Then:
F=(980)(2.8)=2744 N

You might be interested in

The speed at which a light aircraft can take off is 120 km/h. (A) What is the minimum constant acceleration required for the pla

kondor19780726 [428]

Answer:

A) a = 2.31[m/s^2]; B) t = 14.4 [s]

Explanation:

We can solve this problem using the kinematic equations, but firts we must identify the data:

Vf= final velocity = take off velocity = 120[km/h]

Vi= initial velocity = 0, because the plane starts to move from the rest.

dx= distance to run = 240 [m]

$v_{f} ^{2} =v_{i} ^{2}+2*g*dx\\where:\\v_{f}=120[\frac{km}{h} ]*\frac{1hr}{3600sg} * \frac{1000m}{1km} =33.33[m/s]\\\\Replacing\\33.33^{2}=0+2*a*(240)\\ a=\frac{11108.88}{2*240}\\ a=2.31[m/s^2]\\$

To find the time we must use another kinematic equation.

$v_{f} =v_{i} +a*t\\replacing:\\33.33=0+(2.31*t)\\t=\frac{33.33}{2.31}\\ t=14.4[s]$

7 0

3 years ago

You add 500 mL of water at 10°C to 100 mL of water at 70°C. What is the

Sphinxa [80]

Answer:

Option (c) : 20°C

Explanation:

$t(final) = \frac{w1 \times t1 + w2 \times t2}{w1 + w2}$

T(final) = 500* 10 + 100*70/600 = 20°C

4 0

3 years ago

The central star of a planetary nebula emits ultraviolet light with wavelength 104nm. This light passes through a diffraction gr

Gala2k [10]

Answer: 31.33 degrees

Explanation:

The diffraction angles $\theta_{n}$ when we have a slit divided into $n$ parts are obtained by the following equation:

$dsin\theta_{n}=n\lambda$ (1)

Where:

$d$ is the width of the slit

$\lambda$ is the wavelength of the light

$n$ is an integer different from zero.

Now, the first-order diffraction angle is given when $n=1$ , hence equation (1) becomes:

$dsin\theta_{1}=\lambda$ (2)

Now we have to find the value of $\theta_{1}$ :

$sin\theta_{1}=\frac{\lambda}{d}$

$\theta_{1}=arcsin(\frac{\lambda}{d})$ (3)

We know:

$\lambda=104nm=104(10)^{-9}m$

In addition we are told the diffraction grating has 5000 slits per mm, this means:

$d=\frac{1mm}{5000}=\frac{1(10)^{-3}m}{5000}$

Substituting the known values in (3):

$\theta_{1}=arcsin(\frac{104(10)^{-9}m}{\frac{1(10)^{-3}m}{5000}})$

$\theta_{1}=arcsin(0.52)$

<u>Finally:</u>

$\theta_{1}=31.33\º$ >>>This is the first-order diffraction angle

4 0

3 years ago

A 4.0 kg object will have a weight of approximately 14.8 N on Mars. What is the gravitational field strength on M

Pie

Answer:

Gravitational field strength =weight/mass

Explanation:

14.8N/4.0kg

3.7N/kg

3 0

2 years ago

Which is an important step in how an electric motor uses magnetic force to produce motion?

Fiesta28 [93]

Answer:B.As the north pole of the electromagnet nears the south pole of the permanent magnet, the current reverses and the poles of the magnets then repel.

Explanation:

3 0

3 years ago

Read 2 more answers