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3 years ago

Imagine no girls awnser this question ://

Physics

2 answers:

alina1380 [7]3 years ago

4 0

Wdym I am answering rn

Nat2105 [25]3 years ago

3 0

Answer:

im confused

Explanation:

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Richard is driving home to visit his parents. 135{\rm mi} of the trip are on the interstate highway where the speed limit is 65{

nydimaria [60]

Answer:

time spent = 0.2276

Explanation:

given data

distance = 135 mi

usual speed = 65 mph

today speed = 73 mph

solution

we get here time that is express as

time = $\frac{distance}{speed}$ ...................1

usual time = $\frac{135}{65}$ = 2.0769 h

today time = $\frac{135}{73}$ = 1.8493 h

so we get here time spent as

time spent = 2.0769 h - 1.8493 h

time spent = 0.2276

6 0

4 years ago

Tahj walks 28 m south from Imhotep's front doors in order to get into his mothers car. If it takes him 7 s, what was average vel

DochEvi [55]

The correct answer is 4

7 0

3 years ago

Read 2 more answers

A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the

Margaret [11]

Hi there!

$\boxed{\omega = 0.38 rad/sec}$

We can use the conservation of angular momentum to solve.

$\large\boxed{L_i = L_f}$

Recall the equation for angular momentum:

$L = I\omega$

We can begin by writing out the scenario as a conservation of angular momentum:

$I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)$

$I_m$ = moment of inertia of the merry-go-round (kgm²)

$\omega_m$ = angular velocity of merry go round (rad/sec)

$\omega_f$ = final angular velocity of COMBINED objects (rad/sec)

$I_b$ = moment of inertia of boy (kgm²)

$\omega_b$ = angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

$I = MR^2$

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

$\omega = \frac{v}{r}$

$L_b = MR^2(\frac{v}{R}) = MRv$

Plug in the given values:

$L_b = (20)(3)(5) = 300 kgm^2/s$

Now, we must solve for the boy's moment of inertia:

$I = MR^2\\I = 20(3^2) = 180 kgm^2$

Use the above equation for conservation of momentum:

$600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}$

8 0

3 years ago

A velocity-time graph shows how what changes over time.

Mnenie [13.5K]

Answer:

velocity changes over time.

6 0

3 years ago

Compare and contrast electric potential energy and electric potential difference? Explain.

forsale [732]

Answer:

Electric Potential Energy:

The energy that is needed to move a charge against an electric firld is called Electric Potential Energy

Electric Potential Difference:

The amount of work done in carrying a unit charge from one point to an other in an electric field is called Electric Potential Difference.

Relation:

Relation between Electric potential and electrical potential energy is given by

$\delta V=\frac{PE}{q}$

Here PE represents Electric potential energy

and $\delta V$ is Electric potential difference

it means electric potential difference is the difference in electric potential energy divided by the charge.

6 0

3 years ago