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maxonik [38]
3 years ago
7

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
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