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ladessa [460]
3 years ago
14

Read the scenario and solve these two problems.

Physics
2 answers:
Burka [1]3 years ago
7 0

Answers:

a) 5400000 J

b) 45.92 m

Explanation:

a) The kinetic energy K of an object is given by:

K=\frac{1}{2}mV^{2}

Where:

m=12000 kg is the mass of the train

V=30 m/s is the speed of the train

Solving the equation:

K=\frac{1}{2}(12000 kg)(30 m/s)^{2}

K=5400000 J This is the train's kinetic energy at its top speed

b) Now, according to the Conservation of Energy Law, the total initial energy is equal to the total final energy:

E_{i}=E_{f}

K_{i}+P_{i}=K_{f}+P_{f}

Where:

K_{i}=5400000 J is the train's initial kinetic energy

P_{i}=0 J is the train's initial potential energy

K_{f}=0 J is the train's final kinetic energy

P_{f}=mgh is the train's final potential energy, where g=9.8 m/s^{2} is the acceleration due gravity and h is the height.

Rewriting the equation with the given values:

5400000 J=(12000 kg)(9.8 m/s^{2})h

Finding h:

h=45.918 m \approx 45.92 m

Crank3 years ago
6 0

Answer:

1 c   2a

Explanation:

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What is the energy of one mole of photons with a wavelength of 418 nm? (h = 6.63 10-34 J . s, N = 6.02 1023.)
FrozenT [24]

Answer:E = 9.54×10-²Joules

Explanation:

Energy of photon E = hc/¶

Where ¶ is the wavelength

Given h = 6.63×10-³⁴ c= H = 6.02×10²³ ¶ = 418nm = 418 ×10^-9

Substituting the values in the formula

E = 6.63×10-³⁴×6.02×10²³/418 ×10^-9

E = 9.54×10-²Joules

8 0
4 years ago
An 85.0-N crate of apples sits at rest on a ramp that runs from the ground to the bed of a truck. The ramp is inclined at 28.0°
irina [24]

Answer:

A  75.1 N and a direction of 152° to the vertical.

B 85.0 N at 0° to the vertical.

Explanation:

A) The interaction partner of this normal force has what magnitude and direction?

The interaction partner of this normal force is the component of the weight of the crate perpendicular to the ramp. <u>It has a magnitude of 85cos28° = 75.1 N and a direction of 180° - 28° = 152° to the vertical(since it is directed downwards perpendicular to the ramp).</u>

B) The normal and frictional forces are perpendicular components of the contact force exerted on the crate by the ramp. What is the magnitude and direction of the contact force?

Since this force has to balance the weight of the crate, its magnitude is 85.0 N. Its direction has to be vertically opposite to that of the weight.

Since the weight is 180° to the vertical (since it is directed downwards), this force is 0° to the vertical.

<u>So, this force has a magnitude of 85.0 N and a direction of 0° to the vertical.</u>

8 0
3 years ago
A student, standing on a scale in an elevator at rest, sees that his weight is 840 N. As the elevator rises, his weight increase
ololo11 [35]

As per FBD while its accelerating upwards

we can say that

F_n - mg = ma

here normal force is given as

F_n = 1050 N

W = 840 N

now mass is given as

m(9.8) = 840

m = 85.7 kg

now we will have

1050 - 840 = 85.7 \times a

a = 2.45 m/s^2

Now while accelerating downwards we can say by FBD

mg - F_n = ma

again plug in all values

840 - 588 = 85.7 \times a

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5 0
4 years ago
A uniform 1.5-kg rod that is 0.80 m long is suspended at rest from the ceiling by two springs, one at each end of the rod. Both
leonid [27]

Answer:

θ=5.65°

Explanation:

Given Data

Mass m=1.5 kg

Length L=0.80 m

First spring constant k₁=35 N/m

Second spring constant k₂=56 N/m

To find

Angle θ

Solution

As the both springs take half load so apply Hooks Law:

Force= Spring Constant ×Spring stretch

F=kx

x=F/k

as

d=x_{1}-x_{2}\\  as \\x=F/k\\so\\d=\frac{F_{1} }{k_{1}} -\frac{F_{2}}{k_{2}}\\ Where \\F=1/2mg\\d=\frac{(1/2)mg}{k_{1}} -\frac{(1/2)mg}{k_{2}}\\ d=\frac{mg}{2}(\frac{1}{k_{1}} -\frac{1}{k_{2}} )\\ And\\Sin\alpha=d/L\\\\alpha =sin^{-1}[\frac{mg}{2L}(1/k_{1}-1/k_{2})]\\\alpha   =sin^{-1}[\frac{(1.5kg)(9.8m/s^{2} )}{2(0.80m)}(1/35Nm-1/56Nm) ]\\\alpha =5.65^{o}

θ=5.65°

5 0
4 years ago
The driver of a pickup truck accelerates from rest to a speed of 37 mi/hr over a horizontal distance of 215 ft with constant acc
Alona [7]

Answer:

Maximum shearing force developed in each of the two pegs during acceleration is 1830 lbf

Explanation:

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As, the acceleration is uniform, therefore we can use Newton's third equation of motion:

2as = V_{f}^{2}-V_{i}^{2}

First convert speed into ft/sec

1 mile/hr = 1.47 ft/sec

therefore,

37 mile/hr = 37 x 1.47 ft/sec

37 mile/hr =  54.39 ft/sec

with initial speed 0 ft/sec (starting from rest), using in equation of motion:

a = [(54.39 ft/sec)² - (0 ft/sec)²]/2(215 ft)

a = 6.88 ft/sec²

Now, the total shear force will be given by Newton's second law of motion:

F = ma

F = (460 lbm +72 lbm)(6.88 ft/sec²)

F = 3660 lbf

Now for the max shear force in each of the two pegs we divide total fore by 2:

Force in each peg = F/2 = (3660 lbf)/2

<u>Force in each peg = 1830 lbf</u>

4 0
3 years ago
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