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chubhunter [2.5K]

3 years ago

12

Select the correct answer. A golfer hits a golf ball with a velocity of 36.0 meters/second at an angle of28.0°. If the hang time

of the golf ball is 3.44 seconds, what is the range ofthe golf ball? 83 meters 87 meters 108 meters 94 meters

1 answer:

lions [1.4K]3 years ago

5 0

Its 83 im pretty sure :)

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A sound wave has a speed of 330m/s and a wavelength of 0.372 m. what is the frequency of the wave?

Alja [10]

Answer:

887.1Hz

Explanation:

Given parameters:

Speed of sound wave = 330m/s

Wavelength = 0.372m

Unknown:

Frequency = ?

Solution:

To solve this problem, we use the expression below:

Speed = Frequency x wavelength

330 = Frequency x 0.372

Frequency = 887.1Hz

5 0

2 years ago

The electric field direction is defined by the direction of the force felt by (select one of the following answers):A. A negativ

steposvetlana [31]

Answer:

B

Explanation:

The formula for the electric field is Force (N)/charge(Coulombs). The electric field direction is defined by the direction of the force felt by a positive charge.

4 0

3 years ago

Will Mark Brainliest if Correct PLZ!!!!! A bullet is shot at some angle above the horizontal at an initial velocity of 87m/s on

qaws [65]

Answer:

≅50°

Explanation:

We have a bullet flying through the air with only gravity pulling it down, so let's use one of our kinematic equations:

Δx=V₀t+at²/2

And since we're using Δx, V₀ should really be the initial velocity in the x-direction. So:

Δx=(V₀cosθ)t+at²/2

Now luckily we are given everything we need to solve (or you found the info before posting here):

Δx=760 m
V₀=87 m/s
t=13.6 s
a=g=-9.8 m/s²; however, at 760 m, the acceleration of the bullet is 0 because it has already hit the ground at this point!

With that we can plug the values in to get:

$760=(87)(cos\theta )(13.6)+\frac{(0)(13.6^{2}) }{2}$

$760=(1183.2)(cos\theta)$

$cos\theta=\frac{760}{1183.2}$

$\theta=cos^{-1}(\frac{760}{1183.2})\approx50^{o}$

3 0

2 years ago

A fisherman is fishing from a bridge and is using a "44.0-N test line." In other words, the line will sustain a maximum force of

vitfil [10]

Answer:

a) 4.485 kg b) 3.94 kg

Explanation:

since the maximum tension the line can stand is 44 N and for question a the speed is constant (acceleration must be zero since the velocity or speed is not changing), F(tension) = mass * acceleration due to gravity (g) .

44 = m * 9.81m/s^2

m = 44/9.81 = 4.485kg

b) F(tension) = ma + mg ( where a is the acceleration of the body and g is the acceleration of the gravity)

44 = m (a +g)

44 = m (1.37 + 9.81)

44/11.18 = m

m = 3.94 kg

3 0

3 years ago

A ball is dropped from rest at point O. After falling for some time, it passes by a window of height 3.3 m and it does so in 0.2

stiv31 [10]

Answer:

Speed at which the ball passes the window’s top = 10.89 m/s

Explanation:

Height of window = 3.3 m

Time took to cover window = 0.27 s

Initial velocity, u = 0m/s

We have equation of motion s = ut + 0.5at²

For the top of window (position A)

$s_A=0\times t_A+0.5\times 9.81t_A^2\\\\s_A=4.905t_A^2$

For the bottom of window (position B)

$s_B=0\times t_B+0.5\times 9.81t_B^2\\\\s_A=4.905t_B^2$

$\texttt{Height of window=}s_B-s_A=3.3\\\\4.905t_B^2-4.905t_A^2=3.3\\\\t_B^2-t_A^2=0.673$

We also have

$t_B-t_A=0.27$

Solving

$t_B=0.27+t_A\\\\(0.27+t_A)^2-t_A^2=0.673\\\\t_A^2+0.54t_A+0.0729-t_A^2=0.673\\\\t_A=1.11s\\\\t_B=0.27+1.11=1.38s$

So after 1.11 seconds ball reaches at top of window,

We have equation of motion v = u + at

$v_A=0+9.81\times 1.11=10.89m/s$

Speed at which the ball passes the window’s top = 10.89 m/s

7 0

3 years ago