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3 years ago

Which statement is correct?

Physics

1 answer:

NISA [10]3 years ago

5 0

Answer:

1st statement is true

Explanation:

Here statement 1 is correct

Let think about it, if you push down the bar then you are lifting your weight off the pedals.

Obviously, this question does not take into account of racing bikes with straps on pedals, where you would push on one side and pull on the other to match your legs are doing, with straps your other leg can pull pedals upward.

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A 5.0-m radius playground merry-go-round with a moment of inertia of 2000 kg · m 2 is rotating freely with an angular speed of 1

lidiya [134]

Answer:

angular speed = 0.4 rad/s

Explanation:

given data

radius = 5 m

moment of inertia = 2000 kg-m²

angular speed = 1.0 rad/s

mass = 60 kg

to find out

angular speed

solution

Rotational momentum of merry-go-round = I?

we get here momentum that is express as

momentum = 2000 × 1

momentum = 2000 kg-m²/s

and

Inertia of people will be here as

Inertia of people = mr² = 60 × 5²

Inertia of people = 1500 kg-m²

so Inertia of people for two people

1500 × 2 = 3000

and

now conserving angular momentum(ω)

moment of inertia × angular speed = ( momentum + Inertia of people ) angular momentum

2000 × 1 = (2000 + 3000 ) ω

solve we get now

ω = 0.4 rad/s

5 0

2 years ago

The purpose of this lab is to explore the various ways to calculate projectile velocity using horizontal, vertical and angle inf

kolezko [41]

Answer: A projectile is any object in which the only force is gravity

Explanation: Equations on how to calculate projectile velocity is stated below:

The initial velocity Vo being a vector quantity, has two componentsVox and Voy

V0x = V0 cos(θ)

V0y = V0 sin(θ)

The acceleration A is a also a vector with two components Axand Ay given

Ax = 0 and Ay = - g = - 9.8 m/s2

Along the x axis the acceleration is equal to 0 and therefore the velocity Vx is constant

Vx = Vocos(θ)

Along the y axis, the acceleration is uniform and equal to - g and the velocity at time t is g

Vy = Vo sin(θ) - g t

Along the x axis the velocity Vx is constant and therefore the component x of the displacement is

x = Vocos(θ) t

Along the y axis, the motion is of uniform acceleration and the y component of the displacement is

y = Vo sin(θ) t - (1/2) g t2

3 0

3 years ago

The origin of an x axis is placed at the center of a nonconducting solid sphere of radius R that carries a charge +qsphere distr

MA_775_DIABLO [31]

Answer:

$q=49Q/64$

and

$x =16R/15$

Explanation:

See attached figure.

$E_{Q}=$ E due to sphere

$E_{q}=$ E due to particule

$E_{total}=E_{Q}-E_{q}=0$ (1)

according to the law of gauss and superposition Law:

$E_{Q}=E_{1}+E_{2}=E_{2}$ ; electric field due to the small sphere with r1=R/4

$E_{Q}=kq_{2}/(r_{1}^{2})=$

$q_{2}=density*4/3*pi*r_{1}^{3}=Q/(4/3*pi*R^{3})*4/3*pi*r_{1}^{3}=Q*r_{1}^{3}/R^{3}$

then: $E_{Q}=kq_{2}/(r_{1}^{2})=k*Q*r_{1}^{3}/(R^{3}*r_{1}^{2}) = kQ/(4*R^{2})$ (2)

on the other hand, for the particule:

$E_{q}=kq/(r_{p}^{2})$

$r_{p}=2R-R/4=7R/4$ ⇒ $E_{q}=16kq/(49R^{2})$ (3)

We replace (2) y (3) in (1):

$E_{total}=E_{Q}-E_{q}=0=kQ/(4*R^{2}) - 49kq/(16R^{2})$

$q=49Q/64$

--------------------

if R<x<2R AND $E_{total}=E_{Q}-E_{q}=0$

$E_{total}=E_{Q}-E_{q}=0=kQ/(x^{2}) - kq/(2R-x^{2})$

remember that $q=49Q/64$

then:

$Q(2R-x^{2})=49/64*x^{2}$

solving:

$x_{1} =16R/15$

$x_{2} =16R$

but: R<x<2R

so : $x =16R/15$

7 0

2 years ago

HELP PLZ 50 POINTS SPACE QUESTION

Naddik [55]

During the winter, the Northern Hemisphere tilted away from the sun, receiving solar radiation at more of an angle. <u>This results in colder temperatures and more extreme temperature changes.</u>

5 0

2 years ago

The pulse site located at the point where the upper leg bends is called the

VLD [36.1K]

The pulse site located at the point where the upper leg bends is called the femoral. It is an artery found in the thigh. It is large and is deemed as the main arterial supply for the lower part of the body. It is known as the second artery that is the largest. It is being used as the catheter access artery. From it, diagnostics for the heart, brain, arms, kidney and other parts can be directed to the other arterial system. It can also be used as a source to draw blood that is from the arteries when there is low blood pressure.

3 0

3 years ago