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Arisa [49]
2 years ago
11

Why are we unable to work long without food

Physics
1 answer:
nataly862011 [7]2 years ago
7 0

Answer:

Without food your not able to produce energy thats why you cannot work so long

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Susan and Hannah are each riding a swing. Susan has a mass of 25 kilograms, and Hannah has a mass of 30 kilograms. Susan’s swing
Charra [1.4K]

Answer:

Kinetic energy is given by:

K.E. = 0.5 m v²

Susan has mass, m = 25 kg

Velocity with which Susan moves is, v = 10 m/s

Hannah has mass, m' = 30 kg

Velocity with which Hannah moves is, v' = 8.5 m/s

<u>Kinetic energy of Susan:</u>

0.5 m v² = 0.5 × 25 kg × (10 m/s)² = 1250 J

<u>Kinetic energy of Hannah:</u>

0.5 m v'² = 0.5 × 30 kg × (8.5 m/s)² = 1083.75 J

Susan's kinetic energy is <u>1250 J </u>and Hannah's kinetic energy is <u>1083.75 J</u>.

Since kinetic energy is dependent on mass and square of speed. Thus, speed has a greater effect than mass. As it is evident from the above example. Susan has greater kinetic energy due to higher speed than Hannah.

4 0
2 years ago
A ball is dropped from the top of a cliff. By the time it reaches the ground, all the energy in its gravitational potential ener
Bingel [31]

The ball was dropped from a height 20 meters

Explanation:

The given is

1. A ball is dropped from the top of a cliff

2. By the time it reaches the ground, all the energy in its gravitational

   potential energy store has been transferred into its kinetic energy

   store, that mean K.E = P.E

3. The ball is travelling at 20 m/s when it hits the ground

4. The gravitational field strength is 10 N/kg

We need to find the height that the ball dropped from it

The ball dropped from the top of a cliff means the initial speed is 0

→ K.E = \frac{1}{2}m(v^{2}-v_{0}^{2})

where v is the final speed, v_{0} in the initial speed and m

is the mass

→ v = 20 m/s and v_{0} = 0 m/s

→ K.E = \frac{1}{2}m(20^{2}-0^{2})

→ K.E = \frac{1}{2}m(400)

→ K.E = 200 m joules ⇒ when the ball hits the ground

→ P.E = m g h

where g is the gravitational field strength, m is the mass and h is

the height

→ g = 10 N/kg

→ P.E = m(10)(h)

→ P.E = 10 m h joules

→ P.E = K.E

→ 10 m h = 200 m

Divide both sides by 10 m

→ h = 20 meters

The ball was dropped from a height 20 meters

Learn more

You can learn more about gravitational potential energy in brainly.com/question/1198647

#LearnwithBrainly

8 0
2 years ago
A 54 kg man holding a 0.65 kg ball stands on a frozen pond next to a wall. He throws the ball at the wall with a speed of 12.1 m
Inessa [10]

Answer:

The velocity of the man is 0.144 m/s

Explanation:

This is a case of conservation of momentum.

The momentum of the moving ball before it was caught must equal the momentum of the man and the ball after he catches the ball.

Mass of ball = 0.65 kg

Mass of the man = 54 kg

Velocity of the ball = 12.1 m/s

Before collision, momentum of the ball = mass x velocity

= 0.65 x 12.1 = 7.865 kg-m/s

After collision the momentum of the man and ball system is

(0.65 + 54)Vf = 54.65Vf

Where Vf is their final common velocity.

Equating the initial and final momentum,

7.865 = 54.65Vf

Vf = 7.865/54.65 = 0.144 m/s

4 0
2 years ago
If the mass of the block is too large and the block is too close to the left end of the bar (near string B) then the horizontal
iVinArrow [24]

Answer:

xcritical = d− m1 /m2 ( L /2−d)

Explanation: the precursor to this question will had been this

the precursor to the question can be found online.

ff the mass of the block is too large and the block is too close to the left end of the bar (near string B) then the horizontal bar may become unstable (i.e., the bar may no longer remain horizontal). What is the smallest possible value of x such that the bar remains stable (call it xcritical)

. from the principle of moments which states that sum of clockwise moments must be equal to the sum of anticlockwise moments. aslo sum of upward forces is equal to sum of downward forces

smallest possible value of x such that the bar remains stable (call it xcritical)

∑τA = 0 = m2g(d− xcritical)− m1g( −d)

xcritical = d− m1 /m2 ( L /2−d)

6 0
2 years ago
You throw a ball of mass 1 kg straight up. You observe that it takes 2.2 s to go up and down, returning to your hand. Assuming w
Elina [12.6K]

Answer:

10.791 m/s

5.93505 m

Explanation:

m = Mass of ball

v_f = Final velocity

v_i = Initial velocity

t_f = Final time

t_i = Initial time

g = Acceleration due to gravity = 9.81 m/s²

From the momentum principle we have

\Delta P=F\Delta t

Force

F=mg

So,

m(v_f-v_i)=mg(t_f-t_i)\\\Rightarrow v_i=v_f-g(t_f-t_i)\\\Rightarrow v_i=0-(-9.81)(1.1-0)\\\Rightarrow v_i=10.791\ m/s

The speed that the ball had just after it left the hand is 10.791 m/s

As the energy of the system is conserved

K_i=U\\\Rightarrow \dfrac{1}{2}mv_i^2=mgh\\\Rightarrow h=\dfrac{v_i^2}{2g}\\\Rightarrow h=\dfrac{10.791^2}{2\times 9.81}\\\Rightarrow h=5.93505\ m

The maximum height above your hand reached by the ball is 5.93505 m

5 0
2 years ago
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