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3 years ago

Help!!!!!!!!!!!!!!!!

Physics

1 answer:

scoray [572]3 years ago

4 0

Answer:

_Sorrrrry_ I can't help u_

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PLEASE HELP!!

pychu [463]

Answer:

Explanation:

because they are pushing a body

8 0

3 years ago

People have proposed driving motors with the earth's magnetic field. This is possible in principle, but the small field means th

fiasKO [112]

Answer:

636.619772368 A

Explanation:

$\tau$ = Torque = $1\times 10^{-3}\ N/m$

B = Magnetic field of Earth = $5\times 10^{-5}\ T$

A = Area

d = Diameter = 20 cm

Current is given by

$I=\dfrac{\tau}{BA}\\\Rightarrow I=\dfrac{1\times 10^{-3}}{5\times 10^{-5}\times \dfrac{\pi}{4}\times 0.2^2}\\\Rightarrow I=636.619772368\ A$

The current is 636.619772368 A

8 0

3 years ago

During a firework display, a shell is shot through the air with an initiak speed of 70 m/s at an angle of 75° above the horizont

ladessa [460]

Answer:

241.24m

Explanation:

The height at which the shell explodes will be at the maximum height. In projectile motion, maximum height formula is expressed as:

H = u²sin²θ/2g

u is the initial speed = 70m/s

θ the angle of launch = 75°

g is the acceleration due to gravity = 9.81m/s²

Substitute the values into the formula and get H

H = 70²(sin75°)/2(9.81)

H = 4900sin75°/19.62

H = 4900*0.9659/19.62

H = 4733.037/19.62

H = 241.24m

Hence the height at which the shell explodes is 241.24m

4 0

3 years ago

Which of the following variables would you need in order to calculate how long it would take a horizontally launched projectile

Scrat [10]

I’m sorry but I’m going to need a picture.

4 0

3 years ago

A 57 kg skier starts from rest at a height of H = 27 m above the end of the ski-jump ramp. As the skier leaves the ramp, his vel

Hatshy [7]

Answer:

(a) $h=5.95m$

(b) h is the same

Explanation:

According to the law of conservation of energy:

$E_i=E_f\\U_i+K_i=U_f+K_f$

The skier starts from rest, so $K_i=0$ and we choose the zero point of potential energy in the end of the ramp, so $U_f=0$ . We calculate the final speed, that is, the speed when the skier leaves the ramp:

$mgH=\frac{mv^2}{2}\\v=\sqrt{2gH}\\v=\sqrt{2(9.8\frac{m}{s^2})(27m)}\\v=23\frac{m}{s}$

Finally, we calculate the maximum height h above the end of the ramp:

$v_f^2=v_i^2-2gh\\$

The initial vertical speed is given by:

$v_i=vsin\theta$

and the final speed is zero, solving for h:

$h=\frac{v_i^2}{2g}\\h=\frac{((23\frac{m}{s})sin(28^\circ))^2}{2(9.8\frac{m}{s^2})}\\h=5.95m$

(b) We can observe that the height reached does not depend on the mass of the skier

3 0

4 years ago

Help!!!!!!!!!!!!!!!!​

Help!!!!!!!!!!!!!!!!