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3 years ago

If F(theta)=tan theta=3, find F(theta+pi)

Physics

1 answer:

Alex_Xolod [135]3 years ago

4 0

The period of the tan function is π so (∅ + π) would yield the same value as ∅
F(∅ + π) = 3

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Jonathan must lift a 3000n rock using a lever. jonathan uses 300n of force to push down on the lever to move the rock. what is t

-Dominant- [34]

The load is the weight of the rock that Jonathan lifts:
$L=3000 N$
The effort instead is the force applied in input to the lever in order to lift the rock:
$E=300 N$

So, the ratio between load and effort for this exercise is
$\frac{L}{E}= \frac{3000 N}{300 N}=10$
So, the ratio is 10:1.

3 0

3 years ago

(a) At what height above Earth’s surface is the energy required to lift a satellite to that height equal to the kinetic energy r

Nikitich [7]

Answer:

Explanation:

Gravitational Potential Energy at earth surface $U_1=\frac{GM_em}{R_e}$

Gravitational Potential Energy at height h is $U_2=\frac{GM_em}{R_e+h}$

Energy required to lift the satellite $E_1=U_1-U_2$

$E_1=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}$

Now Energy required to orbit around the earth

$E_2=\frac{1}{2}mv_{orbit}^2=\frac{GM_2m}{2(R_e+h)}$

$\Delta E=E_1-E_2$

$\Delta E=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}$

$E_1=E_2$ (given)

$\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}=0$

$\frac{1}{R_e}-\frac{3}{2(R_e+h)}=0$

$h=\frac{R_e}{2}$

$h=3.19\times 10^6\ m$

(b)For greater height $E_1$ is greater than $E_2$

thus energy to lift the satellite is more than orbiting around earth

4 0

3 years ago

Across:

vladimir2022 [97]

Answer:

it would be c i just had it

Explanation:

welcome................

6 0

2 years ago

An 8.75 kg point mass and a 14.0 kg point mass are held in place 50.0 cm apart. A particle of mass m is released from a point be

marysya [2.9K]

Answer

given,

mass of the = m₁ = 8.75 Kg

another mass of the object = m₂ = 14 Kg

distance between them = 50 cm

R₁ = 17 cm

R₂ = 50 -17 = 33 cm

a) Force applied due to the Mass 8.75 in +ve x- direction

$F_1 = \dfrac{GM_1 m}{R_1^2}$

$F_1 = \dfrac{6.67\times 10^{-11} \times 8.75\ m}{0.17^2}$

$F_1 = 2.019\times 10^{8}\ m$

Force applied due to mass 14 Kg in -ve x-direction

$F_2 = \dfrac{GM_2 m}{R_2^2}$

$F_2 = \dfrac{6.67\times 10^{-11} \times 14\ m}{0.33^2}$

$F_2 = 0.857\times 10^{8}\ m$

net force

F = F₁ + F₂

$F = 2.019\times 10^{8}\ m - 0.857\times 10^{8}\ m$

$F = 1.162\times 10^{8}\ m$

Using newton second law

$a = \dfrac{F}{m}$

$a = \dfrac{ 1.162\times 10^{8}\ m}{m}$

$a =1.162\times 10^{8} \ m/s^2$

b) As the acceleration of mass comes out to be +ve hence, the direction will be toward the mass of 8.75 Kg

6 0

3 years ago

How much energy is used for boiling water in a 4 KW kettle for 20 minutes.

Marta_Voda [28]

Answer:About 860,000 results (1.08 seconds)

kettle-power

Jun 18, 2014 — It's interesting to see how much energy heating and cooling systems use ... the kettle uses goes into the water it costs more to use electricity to boil ... 3 minutes is 1/20 of an hour, so 2.4kW for 3 minutes is equivalent to 2.4 / 20 = 0.12kWh. ... kettle should NOT clock over 1kw/h in the 4 minutes it takes to boil

Explanation:

5 0

3 years ago