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3 years ago

Please answer the question in the middle

Physics

1 answer:

Bas_tet [7]3 years ago

5 0

If I remember correctly the second student had the gravity and force and speed of the object to knock down all the pins while the first student didn’t have as much strength if I remember correctly if not I am so sorry

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Which is required for

vaieri [72.5K]

Answer:

You need (B) Two substances in direct contact with one another to have conduction.

I hope this helped at all.

5 0

3 years ago

I NEED HELP SOLVING THIS!!!!!!!!!!!!

Neko [114]

Answer:

Yes

Explanation:

The given parameters are;

The speed with which the fastball is hit, u = 49.1 m/s (109.9 mph)

The angle in which the fastball is hit, θ = 22°

The distance of the field = 96 m (315 ft)

The range of the projectile motion of the fastball is given by the following formula

$Range = \dfrac{u^2 \times sin(2\cdot \theta)}{g}$

Where;

g = The acceleration due to gravity = 9.81 m/s², we have;

$Range = \dfrac{49.1^2 \times sin(2\times22^{\circ})}{9.81} \approx 170.71 \ m$

Yes, given that the ball's range is larger than the extent of the field, the batter is able to safely reach home.

7 0

3 years ago

Suppose you are asked to find the amount of time t, in seconds, it takes for the turntable to reach its final rotational speed.

Annette [7]

Answer:

option (D)

Explanation:

Here initial rotation speed is given, final rotation speed is given and asking for time.

If we use

A) θ=θ0+ω0t+(1/2)αt2

For this equation, we don't have any information about the value of angular displacement and angular acceleration, so it is not useful.

B) ω=ω0+αt

For this equation, we don't have any information about angular acceleration, so it is not useful.

C) ω2=ω02+2α(θ−θ0)

In this equation, time is not included, so it is not useful.

D) So, more information is needed.

Thus, option (D) is true.

5 0

3 years ago

Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. A stationary police car emits a sound of frequen

bixtya [17]

Answer:

a) The velocity of the car is 7.02 m/s and the car is approaching to the police car as the frequency of the police car is increasing.

b) The frequency is 1404.08 Hz

Explanation:

If the police car is a stationary source, the frequency is:

$f_{a} =(\frac{v+v_{c} }{v} )f_{s}$ (eq. 1)

fs = frequency of police car = 1200 Hz

fa = frequency of moving car as listener

v = speed of sound of air

vc = speed of moving car

If the police car is a stationary observer, the frequency is:

$f_{L} =f_{a} (\frac{v}{v-v_{c} } )=(\frac{v+v_{c} }{v-v_{c} } )f_{s}$ (eq. 2)

Now,

fL = frequecy police car receives

fs = frequency police car as observer

a) The velocity of car is from eq. 2:

$1250=1200(\frac{v+v_{c} }{v-v_{c} } )\\1250(v-v_{c} )=1200(v+v_{c} )\\v_{c} =\frac{50*344}{2450} =7.02m/s$

b) Substitute eq. 1 in eq. 2:

$f_{L} =(\frac{v+v_{p} }{v-v_{c} } )(\frac{v+v_{c} }{v-v_{p} } )f_{s} =(\frac{344+20}{344-7.02} )(\frac{344+7.02}{344-20} )*1200=1404.08Hz$

7 0

3 years ago

How fast would you be going (in kmh) if you had a ship that accelerated at a constant 1g for 24 hours?

Nady [450]

Answer:

Explanation:

1 g is 9.8 m/s^2 the problem wants the results in km/h so we'll fix that really quick.

9.8 m/s^2 (1 km/1000m)(60 sec/1 min)^2(60 min/1 hour)^2 = 127008 km/hour^2

Now, I'm assuming the ship is starting from rest, and hopefully you know your physics equations. We are going to use vf = vi + at. Everything is just given, or we can assume, so I'll just solve.

vf = vi + at

vf = 0 + 127008 km/hour^2 * 24 hours

vf = 3,048,192 km/hour

If there's anything that doesn't make sense let me know.

5 0

4 years ago