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3 years ago

Help with 7 , 8 , and 9 anyone?

Physics

1 answer:

makkiz [27]3 years ago

6 0

Seven

Formula

I'm assuming that the formula used is

R = R_ref [1 + α*(T - T_ref)=

Givens

R_ref is the starting resistance of 100 ohms
α (alpha) = 4.0 *10^-4 or 0.00040
T_ref = 11.5 degrees Celcius
T = 10 degrees.
R=??

Solution

R = 100[ 1 + 4.0*10^4(10 - 11.5)] Collapse the brackets.

R = 100[ 1 + 4.0*10^4(-1.5)] Simplify

R = 100[1 - 0.00060] Subtract

R = 100[0.9994] Multiply

R = 99.94

Answer: A

Eight

The accepted formula for resistance and resistivity is

R = ρ * A / L where

rho is the resistivity

A is the face Area of the conductor

L is the length of the conductor.

Now a perfect conductor has a resistance of 0.

So since resistivity is a direct variation with resistance, its value also has to be 0.

Answer: A

Nine

The formula you need for this is W = I^2* R

W = 14 Watts

R = 17 ohms

I = ??

14 = I^2 * 17 Divide bother sides by 17

14/17 = I^2

0.8235 = I^2 Switch and take the square root of both sides.

I = sqrt(0.8234)

I = 0.9075 Answer: D

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svet-max [94.6K]

Answer:

a)The package strikes 256.2 m in the ground relative to the point directly below where it was released

b) The horizontal component will not change it remains same as 41 m/s

c) Vertical component of velocity = 61.41 m/s

Explanation:

a) Consider the vertical motion of plane,

We have equation of motion, s = ut + 0.5 at²

Initial velocity, u = 0 m/s

Displacement, s = 192 m

Acceleration, a = 9.81 m/s²

Substituting

s = ut + 0.5 at²

192 = 0 x t + 0.5 x 9.81 x t²

t = 6.26 seconds

Now we need to find horizontal distance traveled in 6.26 seconds by the package.

We have equation of motion, s = ut + 0.5 at²

Initial velocity, u = 41 m/s

Time, t = 6.26 s

Acceleration, a = 0 m/s²

Substituting

s = ut + 0.5 at²

s = 41 x 6.26 + 0.5 x 0 x 6.26²

s = 256.52 m

The package strikes 256.2 m in the ground relative to the point directly below where it was released

b) The horizontal component will not change it remains same as 41 m/s

c) We have equation of motion, v = u+ at

Initial velocity, u = 0 m/s

Time, t = 6.26 s

Acceleration, a = 9.81 m/s²

Substituting

v = u+ at

v = 0 + 9.81 x 6.26 = 61.41 m/s

Vertical component of velocity = 61.41 m/s

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3 years ago

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Answer:

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Explanation:

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3 years ago

A projectile is launched diagonally into the air and has a hang time of 24.5 seconds. Approximately how much time is required fo

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Answer:

$t=12.25\ seconds$

Explanation:

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It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.

The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is

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Where vo is the magnitude of the initial velocity, $\theta$ is the angle, t is the time and g is the acceleration of gravity

The maximum height the object can reach can be computed as

$\displaystyle t=\frac{v_osin\theta}{g}$

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