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3 years ago

Projectiles Launched Horizontally Quiz

Physics

1 answer:

NikAS [45]3 years ago

7 0

Answer:

3.0 seconds

Explanation:

The time of flight of a projectile (the time it takes to reach the ground) does not depend on the horizontal motion, but only on its vertical motion.

In fact, the time of flight is determined by the suvat equation:

$s=ut+\frac{1}{2}gt^2$

where

s is the vertical displacement

u is the initial vertical velocity (0, in case of these two projectiles)

g = 9.8 m/s^2 is the acceleration of gravity (assuming downward as positive direction)

t is the time of flight

Re-arranging the equation, we get

$t=\sqrt{\frac{2s}{g}}$

We see that this time depends only on s (the heigth of the cliff) and g: therefore, since the two projectiles are launched from the same height, they take the same time to reach the ground, 3.0 seconds.

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Answer:

(1) 1×10⁻⁴

Explanation:

From the question,

α = (ΔL/L)/(ΔT)............. Equation 1

Where α = linear expansivity of the metal plate, ΔL/L = Fractional change in Length, ΔT = Rise in temperature.

Given: ΔL/L = 1×10⁻⁴, ΔT = 10°C

Substitute these values into equation 1

α = 1×10⁻⁴/10

α = 1×10⁻⁵ °C⁻¹ .

β = (ΔA/A)/ΔT................... Equation 2

Where β = Coefficient of Area expansivity, ΔA/A = Fractional change in area.

make ΔA/A the subject of the equation

ΔA/A = β×ΔT.......................... Equation 3

But,

β = 2α.......................... Equation 4

Substitute equation 4 into equation 3

ΔA/A = 2α×ΔT................ Equation 5

Given: ΔT = 5°C, α = 1×10⁻⁵ °C⁻¹

Substitute into equation 5

ΔA/A = ( 2)×(1×10⁻⁵)×(5)

ΔA/A = 10×10⁻⁵

ΔA/A = 1×10⁻⁴

Hence the right option is (1) 1×10⁻⁴

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The electric force between two charged objects of charge +Q0 that are separated by a distance R0 is F0. The charge of one of the

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Answer:

$F=3F_o(\frac{R_o}{R_{o2}})^2$

Explanation:

This problem is approached using Coulomb's law of electrostatic attraction which states that the force F of attraction or repulsion between two point charges, $Q_1$ and $Q_2$ is directly proportional to the product of the charges and inversely proportional to the square of their distance of separation R.

$F=\frac{kQ_1Q_2}{R^2}..................(1)$

where k is the electrostatic constant.

We can make k the subject of formula as follows;

$k=\frac{FR^2}{Q_1Q_2}...........(2)$

Since k is a constant, equation (2) implies that the ratio of the product of the of the force and the distance between two charges to the product of charges is a constant. Hence if we alter the charges or their distance of separation and take the same ratio as stated in equation(2) we will get the same result, which is k.

According to the problem, one of the two identical charges was altered from $Q_o$ to $3Q_o$ and their distance of separation from $R_o$ to $R_{o2}$ , this also made the force between them to change from $F_o$ to $F_{o2}$ . Therefore as stated by equation (2), we can write the following;