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Len [333]
3 years ago
13

A race car has a centriple acceleration of 15.625 m/s as it goes around a curve if the curve is a circle with radius 40 m what’s

is the speed of the car
Physics
1 answer:
Elanso [62]3 years ago
7 0

Answer:

25 m/s

Explanation:

Centripetal acceleration is the square of the tangential velocity divided by the radius.

a = v² / r

15.625 m/s² = v² / (40 m)

v² = 625 m²/s²

v = 25 m/s

The speed of the car is 25 m/s.

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A wave has a wavelength of 4.9 m and a velocity of 9.8 m/s. The medium through which this wave is traveling is then heated so th
garri49 [273]

Answer:

the wavelength is 9.8 meters

Explanation:

We can use the relationship:

Velocity = wavelenght*frequency.

Initially we have:

wavelenght = 4.9m

velocity = 9.8m/s

then:

9.8m/s =  4.9m*f

f = 9.8m/s/4.9m =  2*1/s

now, if the velocity is doubled and the frequency remains the same, we have:

2*9.8m/s = wavelenght*2*1/s

wavelenght = (2*9.8m/s)*(1/2)s = 9.8 m

6 0
3 years ago
Read 2 more answers
On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini
irinina [24]

Answer:

v_a=0.8176 m/s

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being m_a and m_b the masses of pucks a and b respectively, the initial momentum of the system is

M_1=m_av_a+m_bv_b

Since b is initially at rest

M_1=m_av_a

After the collision and being v'_a and v'_b the respective velocities, the total momentum is

M_2=m_av'_a+m_bv'_b

Both momentums are equal, thus

m_av_a=m_av'_a+m_bv'_b

Solving for v_a

v_a=\frac{m_av'_a+m_bv'_b}{m_a}

v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}

v_a=0.8176 m/s

The initial kinetic energy can be found as (provided puck b is at rest)

K_1=\frac{1}{2}m_av_a^2

K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J

The final kinetic energy is

K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2

K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J

The change of kinetic energy is

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

3 0
2 years ago
A radar antenna is tracking a satellite orbiting the earth. At a certain time, the radar screen shows the satellite to be 118 km
ruslelena [56]

Answer:

x component 60.85 m

y component 101.031 m

Explanation:

We have given distance r = 118 km

Angle which makes from ground = 58.9°

(a) X component of distance  is given by r_x=rcos\Theta =118\times cos58.9=118\times 0.5165=118=60.85m

(b) Y component of distance is given by r_Y=rcos\Theta =118\times sin58.9=118\times 0.8562=101.0316m

These are the x and y component of position vector

6 0
3 years ago
What is the frequency of radiation emitted by a photon of light if the energy released during its transition to ground state is
miskamm [114]
If i was feeling harsh today, I'd say the answer to your question is impossible to obtain due to the fact that photons do not emit radiation, photons ARE the radiation emitted. Though for the sake of it, here is the method...

<u>The simple method:
</u>

E=hf

therefore f=e/h

f=(3.611x10^-15) / 6.63x10^-34)

Answer: 5.45x10^18
3 0
3 years ago
A 14.8-g piece of Styrofoam carries a net charge of -0.742 µC and is suspended in equilibrium above the center of a large, horiz
MAXImum [283]
Base on your question where a 14.8g of piece of Styrofoam carries a net charge of -0.742C and is suspended in equilibrium above the center of a large, horizontal sheet of plastic so the ask of the problem is to calculate  the charge per unit area on the plastic sheet. The answer would be 21.96
3 0
3 years ago
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