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2 years ago

6

Kinetic energy is _____.inversely proportional to mass and velocity directly proportional to mass and velocity directly proporti

onal to mass and inversely proportional to velocity inversely proportional to mass and directly proportional to velocity

2 answers:

oee [108]2 years ago

7 0

Answer : The correct option is, directly proportional to mass and velocity.

Explanation:

Kinetic energy : It is associated with the state of motion of an object.

The formula used for kinetic energy is,

$K.E=\frac{1}{2}mv^2$

where,

m = mass of an object

v = velocity or speed of an object

From the formula of kinetic energy we conclude that,

The kinetic energy is directly proportional to the mass and the velocity of an object.

Hence, the kinetic energy is directly proportional to mass and velocity.

saw5 [17]2 years ago

3 0

Directly proportional to mass and velocity

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A projectile has an initial horizontal velocity of 34.0 M/s at the edge of a roof top. Find the horizontal and vertical componen

Sveta_85 [38]

Answer:

$v_x=34 m/s$

$v_y=53.9\ m/s$

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 eventually hits the ground.

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

vx=v

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

$v_y=g.t$

The horizontal component of the velocity is always the same:

$v_x=34 m/s$

The vertical component at t=5.5 s is:

$v_y=9.8*5.5=53.9$

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2 years ago

A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1

bija089 [108]

1.

Answer:

Part a)

$\rho = 1.35 \times 10^{-5}$

Part b)

$\alpha = 1.12 \times 10^{-3}$

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

$i = 18.7 A$

V = 17.0 volts

now we will have

$V = I R$

$17.0 = 18.7 R$

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now we know that

$R = \rho \frac{L}{A}$

$0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho = 1.35 \times 10^{-5}$

Part b)

Now at higher temperature we have

$V = I R$

$17.0 = 17.3 R$

R = 0.98 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho' = 1.46 \times 10^{-5}$

so we will have

$\rho' = \rho(1 + \alpha \Delta T)$

$1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))$

$\alpha = 1.12 \times 10^{-3}$

2.

Answer:

Part a)

$i = 1.55 A$

Part b)

$v_d = 1.4 \times 10^{-4} m/s$

Explanation:

Part a)

As we know that current density is defined as

$j = \frac{i}{A}$

now we have

$i = jA$

Now we have

$j = 1.90 \times 10^6 A/m^2$

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so we will have

$i = 1.55 A$

Part b)

now we have

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so we have

$n = 8.5 \times 10^{28}$

$e = 1.6 \times 10^{-19} C$

so we have

$1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d$

$v_d = 1.4 \times 10^{-4} m/s$

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