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sattari [20]
3 years ago
11

A marble, rolling with speed of 20m/sec rolls off the edge of the table that is 180m high (g=10m/sec2), find time taken to drop

to the ground.
a. 10sec
b. 8sec
c. 6sec
d. 12sec​
Physics
2 answers:
natali 33 [55]3 years ago
6 0

Answer:

<em>Choice: c. 6sec</em>

Explanation:

<u>Horizontal Launch </u>

When an object is thrown horizontally with a speed (v) from a height (h), it describes a curved path ruled by gravity until it finally hits the ground.

The horizontal component of the velocity is always constant because no acceleration exists in that direction, thus:

v_x=v

The vertical component of the velocity changes in time because gravity makes the object fall at increasing speed given by:

v_y=g.t

Where g=10 m/s^2

To calculate the time the object takes to hit the ground, we use the same formula as for free-fall, since the time does not depend on the initial speed:

\displaystyle t=\sqrt{\frac{2h}{g}}

The marble rolls the edge of the table at a height of h=180 m, thus:

\displaystyle t=\sqrt{\frac{2*180}{10}}

\displaystyle t=\sqrt{36}

t = 6 sec

Choice: c. 6sec

Monica [59]3 years ago
5 0

Answer:

The answer is:

C) 6sec

Hope it helps:D

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Answer:

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From energy conservation we know that

\frac{1}{2}mv^2=qV

\frac{1}{2}\times 6.68\times 10^{-27}\times v^2=3.2\times 10^{-19}\times 3.45\times 10^{-3}

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Question 1(Multiple Choice Worth 4 points) The star named Canopus has a declination of approximately –52°. Which of these statem
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Answer:

It is 52° below the celestial equator.

Explanation:

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The star named Canopus has a declination of approximately –52°.

Since the angle is negative, this shows that it is south or below the celestial equator and at 52° south of the celestial equator.

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8 0
3 years ago
A 250. mL sample of gas at 1.00 atm and 20.0°C has the temperature increased to 40.0°C and the volume increased to 500. mL. What
ladessa [460]

Answer:

New pressure is 0.534 atm

Explanation:

Given:

Initial volume of the gas, V₁ = 250 mL

Initial pressure of the gas, P₁ = 1.00 atm

Initial temperature of the gas, T₁ = 20° C = 293 K

Final volume of the gas, V₂ = 500 mL

Final pressure of the gas = P₂

Final temperature of the gas, T₁ = 40° C = 313 K

now,

we know for a gas

PV = nRT

where,

n is the moles

R is the ideal gas constant

also, for a constant gas

we have

(P₁V₁/T₁) = (P₂V₂/T₂)

on substituting the values in the above equation, we get

(1.00 × 250)/293 = (P₂ × 500)/313

or

P₂ = 0.534 atm

Hence, the <u>new pressure is 0.534 atm</u>

5 0
3 years ago
As a stands, her entire weight is momentarily placed on the heels of her high-heeled shoes. Calculate the pressure exerted on th
Lena [83]

Answer:

Pressure will be 6072449.952Pa

Explanation:

We have given mass of the women m = 65 kg

Radius of the heels r = 0.578 cm = 0.00578 m

We have to find the pressure

We know that pressure is given by

P=\frac{F}{A}=\frac{mg}{A}

So force F = mg = 65×9.8 = 637 N

Area A=\pi r^2=3.14\times 0.00578^2=1.049\times 10^{-4}m^2

So pressure p=\frac{637}{1.049\times 10^{-4}}=6072449.952Pa

5 0
3 years ago
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