What is the correct answer?

A foot player runs 1.6m/s and has a KE of 790 J. What is his mass?

Mariana [72]

The equation for kinetic energy is,

Ke = (1/2)mv^2.

You're given a kinetic energy of 790 joules, and a speed of 1.6 m/s. Plugging these values into the equation, we get,

790 = (1/2)(1.6)^2(m).

Solving for m, we get,

m = (790)/(0.5(1.6)^2).

I'll let you crunch out those numbers for yourself :D

If you have any questions, feel free to ask. Hope this helps!

3 0

3 years ago

3. Write any two types of force.

Lerok [7]

Answer:

An <u>applied force</u> is a force that is applied to an object by a person or another object. If a person is pushing a desk across the room, then there is an applied force acting upon the object. The applied force is the force exerted on the desk by the person.

A <u>friction force</u> is the force exerted by a surface as an object moves across it or makes an effort to move across it. There are at least two types of friction force - sliding and static friction. Though it is not always the case, the friction force often opposes the motion of an object. For example, if a book slides across the surface of a desk, then the desk exerts a friction force in the opposite direction of its motion. Friction results from the two surfaces being pressed together closely, causing intermolecular attractive forces between molecules of different surfaces. As such, friction depends upon the nature of the two surfaces and upon the degree to which they are pressed together. The maximum amount of friction force that a surface can exert upon an object can be calculated using the formula below:

$F_{frict}$ = µ • $F_{norm}$

8 0

3 years ago

A 0.750kg block is attached to a spring with spring constant 13.5N/m . While the block is sitting at rest, a student hits it wit

vazorg [7]

Answer:

A)A=0.075 m

B)v= 0.21 m/s

Explanation:

Given that

m = 0.75 kg

K= 13.5 N

The natural frequency of the block given as

$\omega =\sqrt{\dfrac{K}{m}}$

The maximum speed v given as

$v=\omega A$

A=Amplitude

$v=\sqrt{\dfrac{K}{m}}\times A$

$0.32=\sqrt{\dfrac{13.5}{0.75}}\times A$

A=0.075 m

A= 0.75 cm

The speed at distance x

$v=\omega \sqrt{A^2-x^2}$

$v=\sqrt{\dfrac{K}{m}}\times \sqrt{A^2-x^2}$

$v=\sqrt{\dfrac{13.5}{0.75}}\times \sqrt{0.075^2-(0.075\times 0.75)^2}$

v= 0.21 m/s

5 0

3 years ago

in a car moving at constant acceleration, you travel 230 mm between the instants at which the speedometer reads 40 km/hkm/h and

Ronch [10]

The acceleration of the car is 0.8049 $m/s^{2}$ .It takes 13.802s to travel the 230 m.

<h3>What is acceleration?</h3>

In mechanics, acceleration refers to the rate at which an object's velocity with respect to time varies. Acceleration is a vector quantity (in that they have magnitude and direction). The direction of an object's acceleration is determined by the direction of the net force acting on it. Newton's Second Law states that the combined effect of two factors determines how much an item accelerates:

(i) It follows that the magnitude of the net balance of all external forces acting on the object is directly proportional to the magnitude of this net resulting force, and

(ii) the mass of the thing, depending on the materials out of which it is constructed, is inversely proportional to the mass of the thing.

Calculations:

40 km/hr ----- 11.11m/s

80 km/hr ----- 22.22m/s

$v^{2} -u^{2} =2as\\22.22^{2} - 11.11^{2} = 2 x a x 230\\ 370.296=460a\\ a= 0.8049m/s^{2} \\$

Time taken

v-u=at

22.22-11.11= 0.8049 x t

t=13.802s

To learn more about acceleration ,visit:

brainly.com/question/2303856

#SPJ4

4 0

2 years ago

A car accelerates from 13 m/s to 25 m/s in 5.0 s. assume constant acceleration. what was its acceleration?

natima [27]

<span>a = 25-13/6 = 12/6 = 2 m/s^2
Av speed: 25+13/2 = 38/2 = 19 m/sec
Dist = speed * time
19 * 6 = 114 meters</span>

8 0

3 years ago