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igor_vitrenko [27]
2 years ago
8

A catapult can project a projectile at a speed as high as 37.0 m/s. (a) if air resistance can be ignored, how high (in m) would

a projectile launched at this speed rise if projected straight up? m
Physics
1 answer:
motikmotik2 years ago
4 0

Then the maximum height of the catapult will be 69.78 meters.

<h3>What is kinematics?</h3>

The study of motion without considering the mass and the cause of the motion. The equation of motion is given below.

v = u + at

s = ut + (1/2)at²

v² = u² + 2as

A catapult can project a projectile at a speed as high as 37.0 m/s. (a) if air resistance can be ignored.

a = - 9.81 m/s²

u = 37 m/s

v = 0

s = h

Then the maximum height of the catapult will be given by the third equation.

v² = u² + 2as

0² = 37² - 2 × 9.81 × h

h = 37² / (2 × 9.81)

h = 69.78 meters

Then the maximum height of the catapult will be 69.78 meters.

More about the kinematics link is given below.

brainly.com/question/7590442

#SPJ1

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(TCO 4) A signal consists of only two sinusoids, one of 65 Hz and one of 95 Hz. This signal is sampled at a rate of 245 Hz. Find
storchak [24]

Answer:

65 Hz, 95 Hz, 150 Hz, 180 Hz, 310 Hz, 340 Hz

Explanation:

Given :

Frequencies of the sinusoids,

$f_{m_1}= 65 \ Hz$ ,  and

$f_{m_2}= 95 \ Hz$

Sampling rate f_s = \ 245 \ Hz

The positive frequencies at the output of the sampling system are :

$f_{o_1}=\pm f_{m_1} \pm nf_s, f_{o_2}=\pm f_{m_2} \pm nf_s $

When n = 0,

$f_{o_1}= f_{m_1} = 65 \ Hz,\ \  f_{o_2}= f_{m_2} = 95 \ Hz $

when n  = 1,

$f_{o_1}=\pm f_{m_1} \pm f_s, \ \ f_{o_2}=\pm f_{m_2} \pm  f_s $

$f_{o_1}= \pm 65 \pm 245,\ \  f_{o_2}=\pm 95 \pm 245$

$f_{o_1}= 180 \ Hz, 310 \ Hz,\ \  f_{o_2}= 150 \ Hz,340 \ Hz$

When n = 2,

$f_{o_1}= \pm 65 \pm 2(245),\ \  f_{o_2}=\pm 95 \pm 2(245)$

$f_{o_1}= 555 \ Hz, 425 \ Hz,\ \  f_{o_2}= 395 \ Hz,585 \ Hz$

Therefore, the first six positive frequencies present in the replicated spectrum are :

65 Hz, 95 Hz, 150 Hz, 180 Hz, 310 Hz, 340 Hz

8 0
3 years ago
If you and a friend are standing at opposites ends of a gymnasium and one of you claps, will the other hear the clap at the same
34kurt

Answer:

No. Your friend will not hear the clap when he/she sees it.

Explanation:

It takes time for sound waves to go through a large area. It takes longer if people are in the area rather than it is empty.

3 0
3 years ago
At what distance above the surface of the earth is the acceleration due to the earth's gravity 0.855 m/s2 if the acceleration du
natulia [17]
Given: Normal pull of gravity g = 9.8 m/s²; 

 g = 0.855 m/s²  (at a certain distance)

Universal gravitational constant G = 6.67 x 10⁻¹¹ N.m²/Kg²

Mass of the Earth Me = 5.98 x 10²⁴ Kg

Radius r = ?

g = GMe/r²

r = √GMe/g

r = √(6.67 x 10⁻¹¹ N.m²/Kg²)(5.98 x 10²⁴ Kg)/(0.855 m/s²)

r = 2.16 x 10⁷ m or 

r =  21,610 Km





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5 0
3 years ago
Help for physical science u4 limiting reactants
Yuliya22 [10]
The reactants are on the left and the products are on the right of the equation
3 0
4 years ago
For light of wavelength 589 nm, calculate the critical angles for the following substances when surrounded by air.
NeX [460]

(1) The critical angles for the fused quartz is 44⁰,

(2) The critical angles for the polystyrene is 39⁰.

(3) The critical angles for the sodium chloride is 41⁰.

<h3>What is Critical angles?</h3>

In optics, the critical angle is the topmost angle at which a shaft of light traveling in one transparent medium can strike the boundary between that medium and another with a lower refractive indicator without being fully reflected within the first

Critical angles for the different medium.

θ=sin^{-1}\frac{n_{2} }{n_{1} }

where;

n_{2} is the refractive index of air = 1

n_{1} is the refractive index of the given substances.

Refractive index of quartz glass = 1.46

Refractive index of polystyrene = 1.59

Refractive index of sodium chloride = 1.54

Critical angles for quartz glass:

$\theta_{\text {crit }}=\sin ^{-1} \frac{1.00029}{1.46}\\$

$\theta_{\text {crit }}=44^{o}$

Critical angles for polystyrene

$\theta_{\text {crit }}=\sin ^{-1} \frac{1.00029}{1.59}\\\\

\theta_{\text {crit }}=39^{\circ}

Critical angles for Sodium Chloride

$\theta_{\text {crit }}=\sin ^{-1} \frac{1.00029}{1.54}\\

\theta_{\text {crit }}=41^{o}

Thus, the critical angles for the quartz glass is 44⁰, the critical angles for the polystyrene is 39 ⁰ and the critical angles for the sodium chloride is 41⁰.

To know more about critical angle visit:
brainly.com/question/3314727

#SPJ4

I understand that the question you are looking for is:

"For light of wavelength 589 nm, calculate the critical angles for the following substances when surrounded by air.

(a) Fused Quartz

(b) Polystyrene

(c) Sodium chloride"

8 0
2 years ago
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