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3 years ago

A 63.0-kg sprinter starts a race with an acceleration of 4.20 m/s2 . what is the net external force on him?

1 answer:

8 0

From Newton's second law of motion, it was stated that the net external force on a certain object can be calculated by multiplying its mass to the acceleration. This can mathematically be expressed as,
Fnet = m x a
where m is mass and a is acceleration.

Substitute the known values to the equation,
Fnet = (63 kg) x (4.20 m/s²)
Fnet = 264.6 kg m/s²

ANSWER: Fnet = 264.6 N

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A car mass of 1.2 x 10 kilograms starts from rest and attains a speed of 20 meters/seconds in 5 seconds. What net force acted on

lorasvet [3.4K]

Net force on the car=F=4.8 x 10³ N

Explanation:

mass of car= 1.2 x 10³ Kg

initial velocity= Vi=0

Final velocity= Vf= 20 m/s

time = t= 5 s

Using kinematic equation,

Vf= Vi + at

20= 0 + a (5)

5 a=20

a= 20/5

a= 4 m/s²

Now force is given by F = ma

F= 1.2 x 10³ (4)

F=4.8 x 10³ N

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3 years ago

Which of the following is a characteristic of electromagnetic waves? (2 points)

hjlf

What’s the answer choices to pick from?

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4 years ago

An electromagnet is a device in which moving electric charges (current) in a coil of wire create a magnet. What’s one advantage

yan [13]

I think the correct answer from the choices listed above is option D. One advantage of using electromagnets in devices would be that electromagnets can easily be turned on and off. Hope this answers the question. Have a nice day.

7 0

3 years ago

Read 2 more answers

A projectile is launched diagonally into the air and has a hang time of 24.5 seconds. Approximately how much time is required fo

Rasek [7]

Answer:

$t=12.25\ seconds$

Explanation:

Diagonal Launch

It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.

The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is

$x=v_ocos\theta t$

$\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}$

Where vo is the magnitude of the initial velocity, $\theta$ is the angle, t is the time and g is the acceleration of gravity

The maximum height the object can reach can be computed as

$\displaystyle t=\frac{v_osin\theta}{g}$

There are two times where the value of y is $y_o$ when t=0 (at launching time) and when it goes back to the same level. We need to find that time t by making $y=y_o$

$\displaystyle y_o=y_o+v_osin\theta\ t-\frac{gt^2}{2}$

Removing $y_o$ and dividing by t (t different of zero)

$\displaystyle 0=v_osin\theta-\frac{gt}{2}$

Then we find the total flight as

$\displaystyle t=\frac{2v_osin\theta}{g}$

We can easily note the total time (hang time) is twice the maximum (apex) time, so the required time is

$\boxed{t=24.5/2=12.25\ seconds}$

4 0

3 years ago

An airplane accelerates from a speed of 88m/s to a speed of 132 m/s during a 15 second time interval. How far did the airplane t

Gelneren [198K]

Answer:

$1650\:\mathrm{m}$

Explanation:

We can use the following kinematics equations to solve this problem:

$v_f=v_i+at,\\{v_f}^2={v_i}^2+2a\Delta x$ .

Using the first one to solve for acceleration:

$132=88+a(15),\\15a=44,\\a=\frac{44}{15}=2.9\bar{3}\:\mathrm{m/s^2}$ .

Now we can use the second equation to solve for the distance travelled by the airplane:

$132^2=88^2+2\cdot2.9\bar{3}\cdot \Delta x,\\\Delta x= \frac{9680}{2\cdot2.9\bar{3}},\\\Delta x =\fbox{$ 1650\:\mathrm{m}$}$ (three significant figures).

6 0

3 years ago