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3 years ago

Please help meeeeeeeeeeeeeeeee

Physics

1 answer:

Julli [10]3 years ago

5 0

I can't see the answers clearly, but I can see the question. So, I'll just give you a clue/hint.

A stars brightness actually depends on how far it is from your location. If it's far away, it will be dimmer than its counterpart that is closer. To summarize it, if two stars have the same brightness level and one is farther away than the other, the one farther away will appear dimmer than its closer counterpart.

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A cart starts from rest and accelerates at 4.0 m/s2 for 5.0 s, then maintains that velocity for 10 s, and then decelerates at th

zhannawk [14.2K]

Answer:

Final speed of car = 12 m/s

Explanation:

We have equation of motion v = u + at, where v is final velocity, u is initial velocity, a is acceleration and t is time.

a) A cart starts from rest and accelerates at 4.0 m/s² for 5.0 s

v = ?

u = 0 m/s

a = 4.0 m/s²

t = 5 s

v = u + at = 0 + 4 x 5 = 20 m/s

b) Then maintains that velocity for 10 s

v = ?

u = 20 m/s

a = 0 m/s²

t = 10 s

v = u + at = 20 + 0 x 10 = 20 m/s

c) Then decelerates at the rate of 2.0 m/s² for 4.0 s

v = ?

u = 20 m/s

a = -2.0 m/s²

t = 4 s

v = u + at = 20 + -2 x 4 = 12 m/s

Final speed of car = 12 m/s

3 0

3 years ago

A point charge Q is located a distance d away from the center of a very long charged wire. The wire has length L >> d and

zzz [600]

Answer:

$F = \frac{Qq}{2\pi \epsilon_0 L d}$

Explanation:

As we know that if a charge q is distributed uniformly on the line then its linear charge density is given by

$\lambda = \frac{q}{L}$

now the electric field due to long line charge at a distance d from it is given as

$E = \frac{2k\lambda}{d}$

$E = \frac{q}{2\pi \epsilon_0 d}$

now the force on the other charge in this electric field is given as

$F = QE$

$F = \frac{Qq}{2\pi \epsilon_0 L d}$

5 0

4 years ago

Which statement is

ELEN [110]

Answer:

Explanation: This is the correct answer

8 0

3 years ago

Read 2 more answers

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

4 years ago

The differential equation below models the temperature of an 88°C cup of coffee in a 24°C room, where it is known that the coffe

densk [106]

Answer:

$y = 24+64\cdot e^{-150\cdot t}$

Explanation:

Let solve the differential equation by separating corresponding variables:

$\int\limits^t_0\, dt = -\frac{1}{150} \int\limits^y_{y_{o}} \frac{dy}{y-24}$

The solution of this equation is:

$t = -\frac{1}{150}\cdot (\ln|y-24|-\ln |y_{o}-24|)$

The explicit form of the temperature as a function of time is:

$\ln |y-24|=-150\cdot t + \ln |y_{o}-24|$

$y-24 = C\cdot e^{-150\cdot t}$

The value of the integration constant is:

$C = 64$

The complete expression is:

$y = 24+64\cdot e^{-150\cdot t}$

3 0

3 years ago