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lord [1]
2 years ago
6

A car is traveling at the bottom of a 9.00-meter-radius circular hill with a constant speed v. The moment the car is at the bott

om of the hill, it is noted that a person sitting on a scale in the car reads a value off the scale that is 80% more than normally when the scale is at rest in a bathroom. With what speed is the car traveling
Physics
1 answer:
Rina8888 [55]2 years ago
3 0

Answer:

Explanation:

reading of scale = reaction force of surface R

centripetal force = R - mg = m v² / R , m is mass , v is velocity and R is radius of the circular path .

R = mg + m v² / R

given ,

m v² / R = .80 mg

v² = .80 x g x R

= .8 x 9.8 x 9 = 70.56

v = 8.4 m /s

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A student makes the following statement
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The current is not used up. The electrons flow through the entire circuit and this travel is the current. They flow until they are not charged anymore. That is also why the circuit must be closed or else electrons would escape not just light it up for a second then go out.

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What is the wavelength, in nm, of the line in the hydrogen spectrum when one n value is 3 and the other n value is 6?
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Answer:

\lambda=1090nm

Explanation:

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\frac{1}{\lambda}=R_H(\frac{1}{n_1^2}-\frac{1}{n_2^2})

Where R_H is the Rydberg constant for hydrogen and n_1, n_2 are the lower energy state and the higher energy state, respectively.

\frac{1}{\lambda}=1.10*10^{7}m^{-1}(\frac{1}{3^2}-\frac{1}{6^2})\\\frac{1}{\lambda}=9.17*10^{5}m^{-1}\\\lambda=\frac{1}{1.09*10^{6}m^{-1}}\\\lambda=1.09*10^{-6}m*\frac{10^{9}nm}{1m}\\\lambda=1090nm

4 0
3 years ago
A rock is thrown off a 50.0 m high cliff. How fast must the rock leave the cliff top to land on level ground below, 90 m from th
blagie [28]

Answer:

The rock must leave the cliff at a velocity of 28.2 m/s

Explanation:

The position vector of the rock at a time t can be calculated using the following equation:

r = (x0 + v0x · t, y0 + 1/2 · g · t²)

Where:

r = position vector at time t.

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).

Please, see the attached figure for a graphical description of the problem. Notice that the origin of the frame of reference is located at the edge of the cliff so that x0 and y0 = 0.

When the rock reaches the ground, the position vector will be (see r1 in the figure):

r1 = (90 m, -50 m)

Then, using the equation of the vector position written above:

90 m = x0 + v0x · t

-50 m = y0 + 1/2 · g · t²

Since x0 and y0 = 0:

90 m = v0x · t

-50 m = 1/2 · g · t²

Let´s use the equation of the y-component of the vector r1 to find the time it takes the rock to reach the ground and with that time we can calculate v0x:

-50 m = 1/2 · g · t²

-50 m = -1/2 · 9.81 m/s² · t²

-50 m / -1/2 · 9.81 m/s² = t²

t = 3.19 s

Now, using the equation of the x-component of r1:

90 m = v0x · t

90 m = v0x · 3.19 s

v0x = 90 m / 3.19 s

v0x = 28.2 m/s

8 0
3 years ago
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