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zhuklara [117]
3 years ago
9

A student on an amusement park ride moves in circular path with a radius of 3.5 m once every 4.5 s. What is the tangential veloc

ity of the student?
Physics
1 answer:
nata0808 [166]3 years ago
6 0

Answer:

the tangential velocity of the student is 4.89 m/s.

Explanation:

Given;

the radius of the circular path, r = 3.5 m

duration of the motion, t = 4.5 s

let the student's tangential velocity = v

The tangential velocity of the student is calculated as follows;

v = \frac{2\pi r}{t} \\\\v = \frac{2 \pi \times 3.5}{4.5} \\\\v = 4.89 \ m/s

Therefore, the tangential velocity of the student is 4.89 m/s.

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An elevator has a mass of 1000 Kg. What force is needed to accelerate it upward at a rate of 2 m/s/s?
zubka84 [21]

The force needed to accelerate an elevator upward at a rate of 2 m / s^{2} is 2000 N or 2 kN.

<u>Explanation: </u>

As per Newton's second law of motion, an object's acceleration is directly proportional to the external unbalanced force acting on it and inversely proportional to the mass of the object.

As the object given here is an elevator with mass 1000 kg and the acceleration is given as 2 m / s^{2}, the force needed to accelerate it can be obtained by taking the product of mass and acceleration.

                  \text {Force}=\text {Mass} \times \text {Acceleration}

                  \text { Force }=1000 \times 2=2000 \mathrm{N}=2 \text { kilo Newon }

So 2000 N or 2 kN amount of force is needed to accelerate the elevator upward at a rate of 2 m / s^{2}.

3 0
3 years ago
A certain shade of blue has a frequency of 7.15 × 1014 hz. what is the energy of exactly one photon of this light?
Ket [755]
The energy carried by a single photon of frequency f is given by:
E=hf
where h=6.6 \cdot 10^{-34} m^2 kg s^{-1} is the Planck constant. In our problem, the frequency of the photon is f=7.15 \cdot 10^{14}Hz, and by using these numbers we can find the energy of the photon:
E=(6.6\cdot 10^{-34}m^2 kg s^{-1})(7.15 \cdot 10^{14}Hz)=4.7 \cdot 10^{-19}J
4 0
3 years ago
Who was the first to hypothesize that electron orbit a positively charged nucleus
Eva8 [605]
I think the answer is ruthorford
6 0
3 years ago
If a ball dropped from a tower reaches the ground after 3.5 seconds, what is the height of the tower?
solmaris [256]

Answer:

60.025m.

Explanation:

S= ut + at^2/2 (2nd equation of motion)

S= 0 + (9.8)(3.5)^2 /2 (free fall case, initial velocity = 0)

S = 4.9 x 12.25

S= 60.025 m.

Disclaimer: did math in my head, so you better double check with a calculator.

6 0
3 years ago
To get off a frozen lake, a 70 kg person removes his shoe of mass 0.175 kg and throws it horizontally away from the shore at a v
MakcuM [25]

Answer:

the person will be in the shore at 10.73 minutes after launch the shoe.

Explanation:

For this we will use the law of the lineal momentum.

L_i = L_f

Also,

L = MV

where M is de mass and V the velocity.

replacing,

M_i V_i = M_{fp}V_{fp} + M_{fz}V_{fz}

wher Mi y Vi are the initial mass and velocity, Mfp y Vfp are the final mass and velocity of the person and Mfz y Vfz are the final mass and velocity of the shoe.

so, we will take the direction where be launched the shoe as negative. then:

(70)(0) = (70-0.175)(V_fp) + (0.175)(-3.2m/s)

solving for V_fp,

V_fp = \frac{(3.2)(0.175)}{69.825}

V_fp = 0.008m/s

for know when the person will be in the shore we will use the rule of three as:

1 second -------------- 0.008m

t seconds-------------- 5.15m

solving for t,

t = 5.15m/0.008m

t = 643.75 seconds = 10.73 minutes

 

3 0
3 years ago
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