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mixer [17]
3 years ago
12

during a workout, a football player pushes a blocking dummy a distance of 30 m. while pushing the dummy the same distance a seco

nd time, what would the player need to change to be certain that power output increases? the player would need to use a greater force and take more time. the player would need to use a greater force and take less time. the player would need to use a smaller force and take more time. the player would need to use a smaller force and take less time.
Physics
2 answers:
babymother [125]3 years ago
8 0
Power=\frac{Work}{Time}=\frac{Force\times distance}{time}

If he wants to increase power, force must increase and decrease time.
Dmitrij [34]3 years ago
5 0

B.........................

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An object is said to move from a position of 10m East to a position of 5m west. Determine the object's distance travelled.
motikmotik

Answer:

5 i think

Explanation:

4 0
3 years ago
The right ventricle transports oxygenated blood to the lungs.
juin [17]

Answer: False

Explanation: The circulatory system of the body consists of the heart, blood vessels and blood. The deoxygenated blood from the body is carried to the heart.

Here, the deoxygentaed blood is converted into oxygenated by removing carbon dioxide from them and making it oxygenated.

The impure blood from the body is collected by the right ventricle and transported to the lungs for purification and then transported to the body.

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3 years ago
Read 2 more answers
A racing car travels on a circular track of radius 158 m, moving with a constant linear speed of 19.1 m/s. Find its angular spee
SOVA2 [1]

Answer:

\omega=0.12\frac{rad}{s}

Explanation:

In a uniform circular motion, since a complete revolution represents 2π radians, the angular velocity, which is defined as the angle rotated by a unit of time, is given by:

\omega=\frac{2\pi}{T}(1)

Here T is the period, that is, the time taken to complete onee revolution:

T=\frac{2\pi r}{v}(2)

Replacing (2) in (1):

\omega=\frac{2\pi}{\frac{2\pi r}{v}}=\frac{v}{r}\\\omega=\frac{19.1\frac{m}{s}}{158m}\\\omega=0.12\frac{rad}{s}

3 0
3 years ago
Challenge! A marshmallow is dropped from a 5-meter high pedestrian bridge and 0.83 seconds later, it lands right on the head of
WINSTONCH [101]

a₀).  You know ...
         -- the object is dropped from 5 meters
             above the pavement;
         --  it falls for 0.83 second.

a₁).  Without being told, you assume ...
         -- there is no air anyplace where the marshmallow travels,
             so it free-falls, with no air resistance;
         -- the event is happening on Earth,
            where the acceleration of gravity is  9.81 m/s² .

b).  You need to find how much LESS than 5 meters
       the marshmallow falls in 0.83 second.
    
c).  You can use whatever equations you like.
       I'm going to use the equation for the distance an object falls in
       ' T ' seconds, in a place where the acceleration of gravity is ' G '.

d).  To see how this all goes together for the solution, keep reading:


The distance that an object falls in ' T ' seconds
when it's dropped from rest is

                                 (1/2 G) x (T²) .

On Earth, ' G ' is roughly  9.81 m/s², so in 0.83 seconds,
such an object would fall

                               (9.81 / 2) x (0.83)² = 3.38 meters .

It dropped from 5 meters above the pavement, but it
only fell 3.38 meters before something stopped it.
So it must have hit something that was

                         (5.00 - 3.38)  =  1.62 meters

above the pavement.  That's where the head of the unsuspecting
person was as he innocently walked by and got clobbered.

7 0
3 years ago
A force of 14 N acts on a 5 kg object for 3 seconds.
DiKsa [7]

Answer: a) 42Nm b) 8.4m/s

Explanation:

Impulse is defined as object change in momentum.

Since Force = mass × acceleration

F = ma

Acceleration is the rate of change in velocity.

F = m(v-u)/t

Cross multiply

Ft = m(v-u)

Since impulse = Ft

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

The object change in velocity (v-u) = Ft/m from eqn 1

Going to the question;

a) Impulse = Force (F) × time(t)

Given force = 14N and time = 3seconds

Impulse = 14×3

Impulse = 42Nm

b) The object change in velocity (v-u) = Ft/m where mass = 5kg

v-u = 14×3/5

Change in velocity = 42/5 = 8.4m/s

3 0
3 years ago
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