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3 years ago

Wile E. Coyote stands on top of a 90-meter high cliff, looking down at the Roadrunner. If

Physics

1 answer:

Lina20 [59]3 years ago

6 0

Answer:

33.516 kJ

Explanation:

Potential energy is given by:

PE = mgh

Where m is the mass, g is acceleration due to gravity, and h is the height. In this case:

PE = 38kg x 9.8m/s^2 x 90m = 33516 kg m^2/s^2 = 33516 J = 33.516 kJ

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Why is it incorrect to say that astronauts are weightlesS in space while orbiting Earth in a space shuttle?

Mamont248 [21]

This is because Gravity exists everywhere in the universe.

<h3>What is Gravity?</h3>

This is the force of attraction which acts on all matters in the universe. Astronauts appear weightless while in orbiting the Earth because the space shuttle and the astronauts are in free fall around it.

They fall at the same rate as the space shuttle which is why the astronauts appear weightless.

Read more about Gravity here brainly.com/question/88039

6 0

2 years ago

Continuous and aligned fiber-reinforced composite with cross-sectional area of 340 mm2 (0.53 in.2) is subjected to a longitudina

Alecsey [184]

(a) 23.4

The fiber-to-matrix load ratio is given by

$\frac{F_f}{F_m}=\frac{E_f V_f}{E_m V_m}$

where

$E_f = 131 GPa$ is the fiber elasticity module

$E_m = 2.4 GPa$ is the matrix elasticity module

$V_f=0.3$ is the fraction of volume of the fiber

$V_m=0.7$ is the fraction of volume of the matrix

Substituting,

$\frac{F_f}{F_m}=\frac{(131 GPa)(0.3)}{(2.4 GPa)(0.7)}=23.4$ (1)

(b) 44,594 N

The longitudinal load is

F = 46500 N

And it is sum of the loads carried by the fiber phase and the matrix phase:

$F=F_f + F_m$ (2)

We can rewrite (1) as

$F_m = \frac{F_f}{23.4}$

And inserting this into (2):

$F=F_f + \frac{F_f}{23.4}$

Solving the equation, we find the actual load carried by the fiber phase:

$F=F_f (1+\frac{1}{23.4})\\F_f = \frac{F}{1+\frac{1}{23.4}}=\frac{46500 N}{1+\frac{1}{23.4}}=44,594 N$

(c) 1,906 N

Since we know that the longitudinal load is the sum of the loads carried by the fiber phase and the matrix phase:

$F=F_f + F_m$ (2)

Using

F = 46500 N

$F_f = 44594 N$

We can immediately find the actual load carried by the matrix phase:

$F_m = F-F_f = 46,500 N - 44,594 N=1,906 N$

(d) 437 MPa

The cross-sectional area of the fiber phase is

$A_f = A V_f$

where

$A=340 mm^2=340\cdot 10^{-6}m^2$ is the total cross-sectional area

Substituting $V_f=0.3$ , we have

$A_f = (340\cdot 10^{-6} m^2)(0.3)=102\cdot 10^{-6} m^2$

And the magnitude of the stress on the fiber phase is

$\sigma_f = \frac{F_f}{A_f}=\frac{44594 N}{102\cdot 10^{-6} m^2}=4.37\cdot 10^8 Pa = 437 MPa$

(e) 8.0 MPa

The cross-sectional area of the matrix phase is

$A_m = A V_m$

where

$A=340 mm^2=340\cdot 10^{-6}m^2$ is the total cross-sectional area

Substituting $V_m=0.7$ , we have

$A_m = (340\cdot 10^{-6} m^2)(0.7)=238\cdot 10^{-6} m^2$

And the magnitude of the stress on the matrix phase is

$\sigma_m = \frac{F_m}{A_m}=\frac{1906 N}{238\cdot 10^{-6} m^2}=8.0\cdot 10^6 Pa = 8.0 MPa$

(f) $3.34\cdot 10^{-3}$

The longitudinal modulus of elasticity is

$E = E_f V_f + E_m V_m = (131 GPa)(0.3)+(2.4 GPa)(0.7)=41.0 Gpa$

While the total stress experienced by the composite is

$\sigma = \frac{F}{A}=\frac{46500 N}{340\cdot 10^{-6}m^2}=1.37\cdot 10^8 Pa = 0.137 GPa$

So, the strain experienced by the composite is

$\epsilon=\frac{\sigma}{E}=\frac{0.137 GPa}{41.0 GPa}=3.34\cdot 10^{-3}$

3 0

3 years ago

Can a particle with constant speed be accelarating?<br>what if it has constant velocity?

Olin [163]

Explanation:

it can be safely concluded that an object moving in a circle at constant speed is indeed accelerating. It is accelerating because the direction of the velocity vector is changing.

When an object is moving with constant velocity, it does not change direction nor speed and therefore is represented as a straight line when graphed as distance over time.

7 0

3 years ago

Suppose a certain car supplies a constant deceleration of A meter per second per second. If it is traveling at 90km/hr. When. th

aksik [14]

Answer:

i)-6.25m/s

ii)18 metres

iii)26.5 m/s or 95.4 km/hr

Explanation:

Firstly convert 90km/hr to m/s

90 × 1000/3600 = 25m/s

(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)

0 = (25)^2 + 2A(50)

0 = 625 + 100A....then moved the other value to one

-625 = 100A

Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)

(ii) Firstly convert 54km/hr to m/s

In which this is 54 × 1000/3600 = 15m/s

then apply the same formula as that in (i)

0 = (15)^2 + 2(-6.25)s

-225 = -12.5s

Hence the stopping distance = 18metres

(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question

0 = u^2 + 2(-6.25)(56)

u^2 = 700

Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s

In km/hr....26.5 × 3600/1000 = 95.4 km/hr

3 0

3 years ago

HELP PLS

Rama09 [41]

There are a variety of waves from light waves to mechanical waves. Waves can exhibit different effects like the Doppler Effect.

All light waves behave in a similar manner. They either get transmitted, reflected, absorbed, refracted, polarized, diffracted, or scattered based off of the composition of the object and the wavelength of the light.

According to Wikipedia, “One important property of mechanical waves is that their amplitudes are measured in an unusual way, displacement divided by (reduced) wavelength. When this gets comparable to unity, significant nonlinear effects such as harmonic generation may occur, and, if large enough, may result in chaotic effects.” Mechanical waves are chaotic and its “amplitudes” are measured unusually.

Diffraction is when light bends around objects and spread after passing out through small openings. “Diffraction occurs with all waves, including sound waves, water waves, and electromagnetic waves such as light that the eye can see.”-Wikipedia. Here is the formula to Diffraction: <em>d </em>sin <em>θ </em>= <em>nλ</em>

Doppler effect can occur for any type of wave like sound or water waves. An example of this is when we hear a police car with its sirens on, coming towards us. The closer you are to the police car, the higher the wavelength, but the farther away you are, the lower the wavelength.

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5 0

3 years ago