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olasank [31]
1 year ago
10

If a cliff jumper leaps off the edge of a 100m cliff, how long does she fall before hitting the water? (assume zero air resistan

ce)
Physics
1 answer:
andrew-mc [135]1 year ago
4 0
<h2>Answer:</h2>

<em>Hello, </em>

<h3><u>QUESTION)</u></h3>

Assuming that the initial velocity of the jumper is zero, on Earth any freely falling object has an acceleration of 9.8 m/s².  

<em>✔ We have : a = v/Δt = ⇔ Δt = v/a </em>

  • Δt = (√2xgxh)/9,8
  • Δt = (14√10)/9,8
  • Δt ≈ 4,5 s

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A physics professor wants to perform a lecture demonstration of Young's double-slit experiment for her class using the 633-nm li
babunello [35]

Answer:

0.00001266 m

Explanation:

D = Distance from source to screen

m = Order

d = Slit separation

The distance from a point on the screen to the center line

y=\frac{m\lambda D}{d}

At m = 0

y_0=0

y_1-y_0=35\ cm\\\Rightarrow y_1=35\ cm

At m = 1

y_1=\frac{1\times 633\times 10^{-9}\times 7}{d}\\\Rightarrow d=\frac{1\times 633\times 10^{-9}\times 7}{0.35}\\\Rightarrow d=0.00001266\ m

The slit separation is 0.00001266 m

3 0
2 years ago
You decide to work at a heart rate of 150 instead of 120. What area of F.I.T.T. did you change?
Rina8888 [55]

Key concepts

Heart rate

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As Valentine's Day approaches, we're increasingly confronted with "artistic" images of the heart. Real hearts hardly resemble to two-lobed shapes adorning cards and candy boxes this time of year. And the actual shape of the human heart is important for its function of supplying blood to the entire body. You have likely noticed that your heart beats more quickly when you exercise. But have you ever taken the time to observe how long it takes to return to its normal rate after you're done exercising? In this science activity you'll get to do some exercises to explore your own heart-rate recovery time.

Background

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i think it will help you...if it help you ...please mark brainless

8 0
2 years ago
If a 20.0 g object at a temperature of 35.0∘C has a specific heat of 2.89Jg∘C, and it releases 450. J into the atmosphere, what
nataly862011 [7]

Answer:

The final temperature of the object will be 42.785 °C

Explanation:

When the heat added or removed from a substance causes a change in temperature in it, this heat is called sensible heat.

In other words, sensible heat is the amount of heat that a body absorbs or releases without any changes in its physical state (phase change), so that the temperature varies.

The equation for calculating the heat exchanges in this case is:

Q = c * m * ΔT

where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the variation in temperature.

In this case:

  • Q= 450 J
  • c= 2.89 \frac{J}{g*C}
  • m= 20 g
  • ΔT= Tfinal - Tinitial= Tfinal - 35 °C

Replacing:

450 J= 2.89 \frac{J}{g*C} *20 g* (Tfinal - 35°C)

Solving for Tfinal:

\frac{450 J}{2.89\frac{J}{g*C}*20g} =Tfinal -35C

7.785 °C=Tfinal - 35°C

7.785 °C + 35°C= Tfinal

42.785 °C=Tfinal

<u><em>The final temperature of the object will be 42.785 °C</em></u>

8 0
2 years ago
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2 years ago
Daria was swimming in a friend’s pool yesterday, when she saw that a fly had landed in the water about 5 feet away from her. She
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Answer:

Daria probably suffers from Entomophobia.

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