All categories

2 years ago

Please help

Physics

1 answer:

garri49 [273]2 years ago

4 0

Answer:

Exerxingtransforms the chemical energy in the food you eat into Kinetic Energy.

Therefore, the answer is A. Kinetic Energy

You might be interested in

A 30.0-kg child sits on one end of a long uniform beam having a mass of 20.0 kg, and a 40.0-kg child sits on the other end. The

qaws [65]

let the length of the beam be "L"

from the diagram

AD = length of beam = L

AC = CD = AD/2 = L/2

BC = AC - AB = (L/2) - 1.10

BD = AD - AB = L - 1.10

m = mass of beam = 20 kg

m₁ = mass of child on left end = 30 kg

m₂ = mass of child on right end = 40 kg

using equilibrium of torque about B

(m₁ g) (AB) = (mg) (BC) + (m₂ g) (BD)

30 (1.10) = (20) ((L/2) - 1.10) + (40) (L - 1.10)

L = 1.98 m

4 0

2 years ago

A car rounds a flat curve and experiences a centripetal force directed toward the center of the curve and perpendicular to the d

Usimov [2.4K]

<h2>Answer:</h2>

If a car is rounding a flat curve, it experiences a centripetal force that pulls it towards the center of the circle it is rotating in.

Now,

The centripetal force can be balanced by the centrifugal force caused due to the acceleration of the body at the high speed which counters the centripetal force and in turn prevents the car from slipping down the curve.

So,

If the car doesn't hit the gas then the car will fall down from the curve as the Centripetal force will exceed the Centrifugal force of the car.

However, if the car doesn't hit the brake then the car will maintain it's position on the flat curve track as the centrifugal force will counter the effect of centripetal force directed towards the center.

3 0

3 years ago

Read 2 more answers

Let v1, , vk be vectors, and suppose that a point mass of m1, , mk is located at the tip of each vector. The center of mass for

g100num [7]

Answer:

Explanation:

Center of mass is give as

Xcm = (Σmi•xi) / M

Where i= 1,2,3,4.....

M = m1+m2+m3 +....

x is the position of the mass (x, y)

Now,

Given that,

u1 = (−1, 0, 2) (mass 3 kg),

m1 = 3kg and it position x1 = (-1,0,2)

u2 = (2, 1, −3) (mass 1 kg),

m2 = 1kg and it position x2 = (2,1,-3)

u3 = (0, 4, 3) (mass 2 kg),

m3 = 2kg and it position x3 = (0,4,3)

u4 = (5, 2, 0) (mass 5 kg)

m4 = 5kg and it position x4 = (5,2,0)

Now, applying center of mass formula

Xcm = (Σmi•xi) / M

Xcm = (m1•x1+m2•x2+m3•x3+m4•x4) / (m1+m2+m3+m4)

Xcm = [3(-1, 0, 2) +1(2, 1, -3)+2(0, 4, 3)+ 5(5, 2, 0)]/(3 + 1 + 2 + 5)

Xcm = [(-3, 0, 6)+(2, 1, -3)+(0, 8, 6)+(25, 10, 0)] / 11

Xcm = (-3+2+0+25, 0+1+8+10, 6-3+6+0) / 11

Xcm = (24, 19, 9) / 11

Xcm = (2.2, 1.7, 0.8) m

This is the required center of mass

6 0

3 years ago

2. A girl whose mass is 55kg stands on a spring weighing machine inside a lift. When the lift starts to ascend, its acceleration

Annette [7]

Answer:

Explanation:

5 0

3 years ago

What is the function of a synoptic chart

mihalych1998 [28]

Synoptic chart or map is the one that shows the meteorological conditions over an extended region for the particular time period. The other names for synoptic chart are, synoptic scale, large scale or cyclonic scale.

3 0

3 years ago