PHYSICS COLLEGE

As the moon orbits the Earth which of the following changes (1) a. Speed b. Velocity c. Acceleration d. A, B, and C e. None

Answers

Answer 1

Answer:

Answer:

B- Velocity

Explanation:

This means gravity makes the Moon accelerate all the time, even though its speed remains constant.

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An object is constrained by a cord to move in a circular path of radius 1-m on a frictionless, horizontal surface. The cord will break if the tension exceeds 26.9-N. The maximum kinetic energy that this object can have is _____ J. Round your answer to the nearest whole number.

Answers

Answer:

Explanation:

Let m be the mass of the object and v be the maximum velocity . The tension will provide centripetal force for the circular motion .

T = mv² / R where R is radius of circular path . T is tension .

putting the values given in the equation above

26.9 = m v² / 1

m v² = 26.9

kinetic enrgy = 1/2 m v²

= 26.9 / 2

= 13.45 J

13 J .

Maximum kinetic energy = 13 J .

When a hammer thrower releases her ball, she is aiming to maximize the distance from the starting ring. Assume she releases the ball at an angle of 54.6 degrees above horizontal, and the ball travels a total horizontal distance of 30.1 m. What angular velocity must she have achieved (in radians/s) at the moment of the throw, assuming the ball is 1.15 m from the axis of rotation during the spin?

Answers

Answer:

The angular velocity is 15.37 rad/s

Solution:

As per the question:

$\theta = 54.6^(\circ)$

Horizontal distance, x = 30.1 m

Distance of the ball from the rotation axis is its radius, R = 1.15 m

Now,

To calculate the angular velocity:

Linear velocity, v = $\sqrt{(gx)/(sin2\theta)}$

v = $\sqrt{(9.8* 30.1)/(sin2* 54.6)}$

v = $\sqrt{(294.98)/(sin109.2^(\circ))} = 17.67\ m/s$

Now,

The angular velocity can be calculated as:

$v = \omega R$

Thus

$\omega = (v)/(R) = (17.67)/(1.15) = 15.37\ rad/s$

A point charge q1 = 1.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 1 m.(a) Find the electric force on charge q2.
F12 = ? mN
(b) Find the electric force on q1.
F21 = ? mN
(c) What would your answers for Parts (a) and (b) differ if q2 were -6.0 µC?

Answers

To solve this problem we will apply the concepts related to the Electrostatic Force given by Coulomb's law. This force can be mathematically described as

$F = (kq_1q_2)/(d^2)$

Here

k = Coulomb's Constant

$q_(1,2) =$ Charge of each object

d = Distance

Our values are given as,

$q_1 = 1 \mu C$

$q_2 = 6 \mu C$

d = 1 m

$k = 9*10^9 Nm^2/C^2$

a) The electric force on charge $q_2$ is

$F_(12) = ( (9*10^9 Nm^2/C^2)(1*10^(-6) C)(6*10^(-6) C))/((1 m)^2)$

$F_(12) = 54 mN$

Force is positive i.e. repulsive

b) As the force exerted on $q_2$ will be equal to that act on $q_1$ ,

$F_(21) = F_(12)$

$F_(21) = 54 mN$

Force is positive i.e. repulsive

c) If $q_2 = -6 \mu C$ , a negative sign will be introduced into the expression above i.e.

$F_(12) = ((9*10^9 Nm^2/C^2)(1*10^(-6) C)(-6*10^(-6) C))/((1 m)^(2))$

$F_(12) = F_(21) = -54 mN$

Force is negative i.e. attractive

A convex mirror with a focal length of 0.25 m forms a 0.080 m tall image of an automobile at a distance of 0.24 m behind the mirror. What is the magnification of the image? Where is the car located, and what is its height? Is the image real or virtual? Is the image upright or inverted? Draw a ray diagram to show where the image forms and how large it is with respect to the object

Answers

Answer:

The distance and height of the object is 6 m and 2 m.

The image is virtual and upright.

Explanation:

Given that,

Focal length = 0.25 m

Length of image = 0.080 m

Image distance = 0.24 m

We need to calculate the distance of the object

Using formula of lens

$(1)/(v)=(1)/(f)+(1)/(u)$

Put the value into the formula

$(1)/(0.24)=(1)/(0.25)+(1)/(u)$

$(1)/(u)=(1)/(0.24)-(1)/(0.25)$

$(1)/(u)=(1)/(6)$

$u=6\ m$

We need to calculate the magnification

Using formula of magnification

$m=-(v)/(u)$

Put the value into the formula

$m=-(0.24)/(-6)$

$m=0.04$

We need to calculate the height of the object

Using formula of magnification

$m=(h')/(h)$

$h=(0.080)/(0.04)$

$h=2\ m$

A convex mirror produce a virtual and upright image behind the mirror.

Hence, The distance and height of the object is 6 m and 2 m.

The image is virtual and upright.

Answer:

Distance of the object = 6 m

Height of the object = 2 m

Explanation:

Thinking process:

Given that,

Focal length = 0.25 m

Length of image = 0.080 m

Image distance = 0.24 m

We need to calculate the distance of the object

Therefore, using formula of lens:

$(1)/(u) = (1)/(f) + (1)/(u)$

$(1)/(u) = (1)/(6)$

solving, gives u = 6

The magnification is calculated as follows:

m = -0.24/-6

= 0.04

The height = 2 m

The diagram yields an image behind the mirror which is upright.

Which of the following best represents stored potential energy?Air leaking from a flat tire
Stress built up in a rock fault
Heat given off by a forest fire
Water flowing through a hose

Answers

Answer:

Explanation:

stress built up on a rock fault

A squirrel is running a race where she is on track for her average velocity to be 6.0 m/s. She is distracted by a dummy of an attractive male squirrel and pauses for 3.0 s. As a result her average velocity ends up being 5.0 m/s instead. What is the length of the race? [HINT: construct two equations with the same two unknowns in them and you can solve the system of equations]