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3 years ago

7

What happens to the air resistance on a car as it speeds up? A) decreases b)stays the same C)it changes direction of the car D)i

ncreases

1 answer:

Naya [18.7K]3 years ago

5 0

the answer is D- increases

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How does mass affect acceleration during free fall?

nirvana33 [79]

The first one, as the mass is higher so it accelerates more

3 0

3 years ago

The drawing shows a hydraulic chamber with a spring (spring constant = 1600 N/m) attached to the input piston and a rock of mass

Triss [41]

Answer:

$\Delta x=245\ mm$

Explanation:

Given:

spring constant of the spring attached to the input piston, $k=1600\ N.m^{-1}$

mass subjected to the output plunger, $m=40\ kg$

<u>Now, the force due to the mass:</u>

$F=m.g$

$F=40\times 9.8$

$F=392\ N$

<u>Compression in Spring:</u>

$\Delta x=\frac{F}{k}$

$\Delta x=\frac{392}{1600}$

$\Delta x=0.245\ m$

or

$\Delta x=245\ mm$

8 0

4 years ago

maks197457 [2]

Hey there!
Alright, it looks like you've got the second question on the top right, but I can't figure out the answer you've got for the first.
Using Newton's Second Law Of Motion, we have:
F = ma
We plug in our values:
F = 5(1)
That would give us 5 Newtons of force.
For the last one:
Even if two balls are given the same acceleration, the determining factor is <em>weight.</em> If a ball is lighter, it needs less acceleration to go faster. If it's larger, it needs more to make it go at that same speed. That's why that 5 newtons makes sense because it's 5 kilograms as opposed to 0.6 kilograms. It's heavier, and needs more force.
Hope this helped!

8 0

3 years ago

The dimensions of aluminum foil in a box for sale in super markets are 66 2/3 yards by 12 inches. the mass of the foil is 0.83 k

disa [49]

Length of the sheet is given as

$L = \frac{200}{3} yards = 6096 cm$

width of the sheet is given as

$w = 12 inches = 30.48 cm$

now let say its thickness is "t"

so the volume of the sheet is given as

$V = L*w*t$

$V = 6096*30.48* t$

$V = 185806.08*t cm^3$

mass of the sheet is given as

$m = 0.83 kg = 830 gram$

now we have

$density = \frac{mass}{volume}$

$2.70 = \frac{830}{185806.08*t}$

by solving above we have

$t = 1.65 * 10^{-3} cm$

so the thickness of sheet will be above

4 0

3 years ago

Calculate the ratio of the drag force on a jet flying at 950 km/h at an altitude of 10 km to the drag force on a prop-driven tra

Black_prince [1.1K]

Answer:

$\frac{F_1}{F_2}=3.55$

Explanation:

F = Force

C = Drag coefficient equal for both aircrafts

ρ = Density of air

A = Surface area equal for both aircrafts

v = Velocity

$v_2=\frac{2}{5}v_1$

$F_1=\frac{1}{2}\rho_1 CAv_1^2$

$F_2=\frac{1}{2}\rho_2 CAv_2^2\\\Rightarrow F_2=\frac{1}{2}\rho_2 CA\left(\frac{2}{5}v_1\right)^2$

Dividing the above two equations we get

$\frac{F_1}{F_2}=\frac{\frac{1}{2}\rho_1 CAv_1^2}{\frac{1}{2}\rho_2 CA\left(\frac{2}{5}v_1\right)^2}\\\Rightarrow \frac{F_1}{F_2}=\frac{\rho_1}{\rho_2\frac{4}{25}}\\\Rightarrow \frac{F_1}{F_2}=\frac{0.38}{0.67\frac{4}{25}}\\\Rightarrow \frac{F_1}{F_2}=3.55$

The ratio of the drag forces is $\mathbf{\frac{F_1}{F_2}}=\mathbf{3.55}$

5 0

3 years ago