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3 years ago

A baseball is hit nearly straight up into the air with a speed of 22 m/s. (a) how high does it go ?

1 answer:

4 0

So, this is a problem where the accleration is not provided, since it is implied. The only acceleration is acceleration due to gravity (9.8 m/s)

The equation we will use for this problem is $V^2 =V_{0}^2 + 2a (X-X_0)$

V is the final velocity, V₀ is the initial velocity, a is the acceleration, X is the final height, and X₀ is the starting height.

We can assume that the ball starts on the ground since no height is given, so now we plug our numbers in.

We will use 0 as the final velocity, since the ball will stop moving upwards when it is the highest. We will use -9.8 since that is the acceleration due to gravity and we will use 22m/s as V₀ since that is the starting velocity.

$V^2 =V_{0}^2 + 2a (X-X_0)\\0^2 = 22^2 + 2\times-9.8(X-0)\\0=484-19.6x\\-484=-19.6x\\24.69387755 = x\\x\approx24.69$

So, the ball will go 24.69 meters up

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If the Earth were compressed in such a way that its mass remained the same, but the distance around the equator were just one-ha

vovikov84 [41]

If the distance around the equator is reduced by half, then the radius is also reduced by half.

Since the acceleration due to gravity is proportional to 1/(radius²),
the acceleration changes by a factor of 1/(1/2)² = 1/(1/4) = 4 .

The acceleration due to gravity ... and also the weight of everything on Earth ...
becomes 4 times what it is now.

6 0

3 years ago

Read 2 more answers

In an aqueous solution where the H+ concentration is 1 x 10-6 M, the OH concentration must be:

vesna_86 [32]

Answer:

D. 1×10⁻⁸ M

Explanation:

[H⁺] [OH⁻] = 10⁻¹⁴

(1×10⁻⁶) [OH⁻] = 10⁻¹⁴

[OH⁻] = 1×10⁻⁸

6 0

3 years ago

A high-speed flywheel in a motor is spinning at 450 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and

alexira [117]

Answer:

A) $\omega_f=17.503\ rad.s^{-1}$

B) $t=55.6822\ s$

C) $\theta=1312\ rad$

Explanation:

Given:

mass of flywheel, $m=40\ kg$

diameter of flywheel, $d=0.72\ m$

rotational speed of flywheel, $N_i=450\ rpm \Rightarrow \omega_i=\frac{450\times 2\pi}{60} =15\pi\ rad.s^{-1}$

duration for which the power is off, $t_0=35\ s$

no. of revolutions made during the power is off, $\theta=180\times 2\pi=360\pi\ rad$

Using equation of motion:

$\theta=\omega_i.t+\frac{1}{2} \alpha.t^2$

$360\pi=15\pi\times 35+\frac{1}{2} \times \alpha\times35^2$

$\alpha=-0.8463\ rad.s^{-2}$

Negative sign denotes deceleration.

Now using the equation:

$\omega_f=\omega_i+\alpha.t$

$\omega_f=15\pi-0.8463\times 35$

$\omega_f=17.503\ rad.s^{-1}$ is the angular velocity of the flywheel when the power comes back.

Here:

$\omega_f=0\ rad.s^{-1}$

Now using the equation:

$\omega_f=\omega_i+\alpha.t$

$0=15\pi-0.8463\times t$

$t=55.6822\ s$ is the time after which the flywheel stops.

Using the equation of motion:

$\theta=\omega_i.t+\frac{1}{2} \alpha.t^2$

$\theta=15\pi\times 55.68225-0.5\times 0.8463\times 55.68225^2$

$\theta=1312\ rad$ revolutions are made before stopping.

3 0

3 years ago

If the rectangular barge is 3.0 m by 20.0 m and sits 0.70 m deep in the harbor, how deep will it sit in the river?

leva [86]

The harbour contains salt water while the river contains fresh water. So assuming that the densities of fresh water and salt water are:

density (salt water) = 1029 kg / m^3

density (fresh water) = 1000 kg / m^3

The amount of water (in mass) displaced by the barge should be equal in two waters.

mass displaced (salt water) = mass displaced (fresh water)

Since mass is also the product of density and volume, therefore:

[density * volume]_salt water = [density * volume]_fresh water ---> 1

First we calculate the amount of volume displaced in the harbour (salt water):

V = 3.0 m * 20.0 m * 0.70 m

V = 42 m^3 of salt water

Plugging in the values into equation 1:

1029 kg / m^3 * 42 m^3 = 1000 kg/m^3 * Volume fresh water

Volume fresh water displaced = 43.218 m^3

Therefore the depth of the barge in the river is:

43.218 m^3 = 3.0 m * 20.0 m * h

h = 0.72 m (ANSWER)

8 0

3 years ago

11)A 1100 kg car travels on a straight highway with a speed of 30 m/s. The driver sees a red light ahead and applies her

cestrela7 [59]

Answer:

Time taken = 8.25 second

Explanation:

Given:

Force = 4000 N

Force = ma

4,000 = (1100)(a)

Acceleration = 3.6363 m/s²

v = u + at

0 = 30 + (3.6363)t

Time taken = 8.25 second

3 0

3 years ago