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3 years ago

NEED HELP ASAP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Physics

2 answers:

gulaghasi [49]3 years ago

7 0

Answer:

It would be cooler I believe

Explanation:

Delicious77 [7]3 years ago

6 0

Answer:

warmer

Explanation:

I really think it just gets hot

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You stand on a merry-go-round which is spinning at f = 0:25 revolutions per second. You are R = 200 cm from the center. (a) Find

ivanzaharov [21]

Answer:

(a) ω = 1.57 rad/s

(b) ac = 4.92 m/s²

Explanation:

(a)

The angular speed of the merry go-round can be found as follows:

ω = 2πf

where,

ω = angular speed = ?

f = frequency = 0.25 rev/s

Therefore,

ω = (2π)(0.25 rev/s)

ω = 1.57 rad/s

(b)

The centripetal acceleration can be found as:

ac = v²/R

but,

v = Rω

Therefore,

ac = (Rω)²/R

ac = Rω²

therefore,

ac = (2 m)(1.57 rad/s)²

ac = 4.92 m/s²

(c)

In order to avoid slipping the centripetal force must not exceed the frictional force between shoes and floor:

Centripetal Force = Frictional Force

m*ac = μs*R = μs*W

m*ac = μs*mg

ac = μs*g

μs = ac/g

μs = (4.92 m/s²)/(9.8 m/s²)

μs = 0.5

7 0

3 years ago

Let’s say I am in a bumper car and have a velocity of 14 m/s, driving in the positive x-direction. I and my bumped car have a ma

AlekseyPX

Answer:

160 kg

12 m/s

Explanation:

$m_1$ = Mass of first car = 120 kg

$m_2$ = Mass of second car

$u_1$ = Initial Velocity of first car = 14 m/s

$u_2$ = Initial Velocity of second car = 0 m/s

$v_1$ = Final Velocity of first car = -2 m/s

$v_2$ = Final Velocity of second car

For perfectly elastic collision

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow m_2v_2=m_{1}u_{1}+m_{2}u_{2}-m_{1}v_{1}\\\Rightarrow m_2v_2=120\times 14+m_2\times 0-(120\times -2)\\\Rightarrow m_2v_2=1920\\\Rightarrow m_2=\frac{1920}{v_2}$

Applying in the next equation

$v_2=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 120}{120+\frac{1920}{v_2}}\times 14+\frac{m_2-m_1}{m_1+m_2}\times 0\\\Rightarrow \left(120+\frac{1920}{v_2}\right)v_2=3360\\\Rightarrow 120v_2+1920=3360\\\Rightarrow v_2=\frac{3360-1920}{120}\\\Rightarrow v_2=12\ m/s$

$m_2=\frac{1920}{v_2}\\\Rightarrow m_2=\frac{1920}{12}\\\Rightarrow m_2=160\ kg$

Mass of second car = 160 kg

Velocity of second car = 12 m/s

5 0

4 years ago

The flywheel is rotating with an angular velocity ω0 = 2.37 rad/s at time t = 0 when a torque is applied to increase its angular

nika2105 [10]

Answer:

ω = 12.023 rad/s

α = 222.61 rad/s²

Explanation:

We are given;

ω0 = 2.37 rad/s, t = 0 sec

ω =?, t = 0.22 sec

α =?

θ = 57°

From formulas,

Tangential acceleration; a_t = rα

Normal acceleration; a_n = rω²

tan θ = a_t/a_n

Thus; tan θ = rα/rω² = α/ω²

tan θ = α/ω²

α = ω²tan θ

Now, α = dω/dt

So; dω/dt = ω²tan θ

Rearranging, we have;

dω/ω² = dt × tan θ

Integrating both sides, we have;

(ω, ω0)∫dω/ω² = (t, 0)∫dt × tan θ

This gives;

-1[(1/ω_o) - (1/ω)] = t(tan θ)

Thus;

ω = ω_o/(1 - (ω_o × t × tan θ))

While;

α = dω/dt = ((ω_o)²×tan θ)/(1 - (ω_o × t × tan θ))²

Thus, plugging in the relevant values;

ω = 2.37/(1 - (2.37 × 0.22 × tan 57))

ω = 12.023 rad/s

Also;

α = (2.37² × tan 57)/(1 - (2.37 × 0.22 × tan 57))²

α = 8.64926751525/0.03885408979 = 222.61 rad/s²

6 0

3 years ago

Show the weight of the ladder and draw the missing Frictional force.

madreJ [45]

Since there is no friction between the ladder and the wall, there can be no vertical force component. That's the tricky part ;)

So to find the weight, divide the 100N normal force by earths gravitational acceleration, 9.8m/s^2

$W = \frac{N}{g} = \frac{100N}{9.8m/s^{2}} = \frac{100}{9.8} = 10.2kg$

Then;
Draw an arrow at the base of the ladder pointing towards the wall with a value of 30N, to show the frictional force.

5 0

3 years ago

Heptane is always composed of 84.0% carbon and 16.0% hydrogen. This illustrates the law of multiple proportions. conservation of

BigorU [14]

Heptane is always composed of 84.0% carbon and 16.0% hydrogen. This illustrates the "law of definite proportions".

Answer: Option C

Explanation:

Proust's law states that every chemical compound used to made up of element constituents with constant proportions in terms of its mass and also independent from its sources and synthesis method. In 1779, Joseph Proust gave other names to the Proust's law as, the law of composition or definite proportions or constant compositions.

This can understood from given example like: Oxygen is composed of 8/9 of the mass of any sample of pure water while the hydrogen fills up the remaining 1/9 of the mass. The basis of stoichiometry is structured with the law of multiple proportions along the law of definite proportions.

5 0

3 years ago