How long does it take a car to cross a 20m bridge if it starts from rest and accelerates at 5 m/s^2?

A 23.3-kg mass is attached to one end of a horizontal spring, with the other end of the spring fixed to a wall. The mass is pull

sdas [7]

In spring mass system we know that angular frequency is given as

$\omega = 2\pi f$

f = 8.38 Hz

$\omega = 2\pi(8.38)$

$\omega = 52.65 rad/s$

now we know that speed of SHM at its extreme position is given by

$v = A\omega$

here we know that

A = 17.5 cm

$v = 0.175 (52.65)$

$v = 9.21 m/s$

so maximum speed is 9.21 m/s

7 0

3 years ago

A wrestler weighs in for the first match on the moon. will the athlete weigh more or less on the moon than he does on Earth?

Firdavs [7]

The moon is 1/4 the size of Earth, so the moon's gravity is much less than the earth's gravity, 83.3% (or 5/6) less to be exact. Finally, "weight" is a measure of the gravitational pull between two objects. So of course you would weigh much less on the moon.

4 0

3 years ago

What happens if you move a bar magnet back and forth along the axis of the

nadezda [96]

C) A current is induced in the coiled wire, which lights the light bulb

The moving magnetic field creates electricity which lights the light bulb

Hope it helps!

8 0

1 year ago

Read 2 more answers

A 3 kg penguin is pushed by his penguin friends who give him an initial speed vo at the top of a 30 m hill. The penguin is hopin

Strike441 [17]

Answer:

This question can be answered by using conversation of energy.

$K_1 + U_1 = K_2 + U_2$

$\frac{1}{2}mv_{0}^2 + mgh_1 = 0 + mgh_2$

$\frac{1}{2}(3)v_0^2 + (3)(9.8)(30) = (3)(9.8)(45)\\\frac{1}{2}(3)v_0^2 = 441\\v_0^2 = 294\\v_0 = 17.14 m/s$

Explanation:

Note that we take $K_2 = 0$ because we are looking for the minimum initial speed for the penguin to reach the top of the second hill. Any other speed more than this will already be enough for him.

7 0

3 years ago

Six friends are at a pizzeria. They want to order enough pizza so that each person can eat at least 2/5 . start fraction, 2, d

Lyrx [107]

Friend #1 gets at least 2/5 of a pizza.

Friend #2 gets at least 2/5 .

Friend #3 gets at least 2/5 .

Friend #4 gets at least 2/5 .

Friend #5 gets at least 2/5 .

Friend #6 gets at least 2/5 .

Sum . . . . . . . . . at least 12/5 of a pizza.

Simplify . . . . . . at least 2.4 pizzas.

-- If pizzas can be bought by the half, they should order at least 2-1/2 pizzas.

-- If only whole pizzas have to be ordered, then they should order at least 3 pizzas.

4 0

3 years ago