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joja [24]
3 years ago
6

3. The soccer kicker kicks a 2 kg football with a force of 68 N. How fast will the ball accelerate down the field?

Physics
1 answer:
Katen [24]3 years ago
6 0

Answer:

\boxed {\boxed {\sf m= 2 \ kg , \\a= 34 \ m/s^2,  \\\F= 68 \ N}}

Explanation:

The formula for force is:

F=m*a

If we rearrange the formula to solve for a (acceleration), the formula becomes

\frac{F}{m} =a

The force is 68 Newtons. Let's convert the units to make the problem easier later on. 1 N is equal to 1 kg*m/s², so the force of 68 N is equal to 68 kg*m/s².

The mass is 2 kilograms.

F=68 \ kg*m/s^2 \\m= 2\ kg

Substitute the values into the formula.

\frac{68 \ kg*m/s^2}{2 \ kg} =a

Divide. Note that the kilograms will cancel each other out (hence why we changed the units).

\frac{68 \ m/s^2}{2}=a

34 \ m/s^2=a

The acceleration is<u> </u><u>34 meters per second squared.</u>

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Convert time from 12-hour to 24-hour clock. ​
svet-max [94.6K]

Answer:

4:28

Explanation:

4:28am

a 12 hour clock continues going up after 12 (1:00pm=13:00). minutes stay the same. 12:00pm=00:00. this shows 4:28am, so you count 4 after 00:00.

6 0
3 years ago
In the Bohr model of the hydrogen atom, an electron({rm mass};m=9.1; times 10^{ - 31;}{rm kg}) orbits a proton at a distance of
max2010maxim [7]

Answer:

n=6.56×10¹⁵Hz

Explanation:

Given Data

Mass=9.1×10⁻³¹ kg

Radius distance=5.3×10⁻¹¹m

Electric Force=8.2×10⁻⁸N

To find

Revolutions per second

Solution

Let F be the force of attraction

let n  be the number of revolutions per sec made by the electron around the nucleus then the centripetal force is given by

F=mω²r......................where ω=2π  n

F=m4π²n²r...............eq(i)

as the values given where

Mass=9.1×10⁻³¹ kg

Radius distance=5.3×10⁻¹¹m

Electric Force=8.2×10⁻⁸N

we have to find n from eq(i)

n²=F/(m4π²r)

n^{2} =\frac{8.2*10^{-8} }{9.11*10^{-31}* 4\pi^{2} *5.3*10^{-11}  }\\ n^{2}=4.31*10^{31}\\ n=\sqrt{4.31*10^{31}}\\ n=6.56*10^{15}Hz

8 0
3 years ago
A 5 kg block is being pushed horizontally across a level surface at a constant velocity. What is the magnitude of the Normal for
luda_lava [24]

Answer:

50 N.

Explanation:

On top of a horizontal surface, the normal force acting on an object is equivalent to the force of gravity acting on the object. That is:

\displaystyle \begin{aligned} F_N = F_g & = ma \\ & = mg\end{aligned}

The mass of the block is 5 kg and the given force due to gravity is 10 N/kg. Substitute and evaluate:

\displaystyle F_N = F_g = (5\text{ kg})(10 \text{ N/kg}) = 50 \text{ N}

In conclusion, the normal force acting on the block is 50 N.

5 0
2 years ago
Can someone please help me?
svetlana [45]

Answer:

acceleration...............

7 0
3 years ago
Read 2 more answers
A 600‑kg car accelerates at a rate of 3 m/s2. How much net force is acting on the car to cause this acceleration?
finlep [7]
600(kg) x 3(m/s^2) = 1800N (newtons)
4 0
3 years ago
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