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3 years ago

8

A jumbo jet of mass 100,000 kg during takeoff experience a thrust for each of its four engines of 50,000 N. producing an acceler

ation of
2 m/s2
1 m/s
0.5 m/s2
O 4 m/s²
none of the above

1 answer:

PtichkaEL [24]3 years ago

8 0

Answer:

F=MA

5OOOO=100000×A

50000/100000=A

A=5/10 M/ S SQ

A=0.5 M/S SQ.

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A bumper cart has a mass of 200 kg and has a protective bumper around it that behaves like a spring. The spring constant is 5000

34kurt

Part A:
For this part we’re assuming all the kinetic energy of the moving bumper car is converted into elastic potential energy in the spring since the car is brought to rest. Therefore you can find the total kinetic energy to get your answer:

KE = ½ mv^2
KE = ½ (200)(8)^2
KE = 6400 J

Part B:
Now you can use Hooke’s law to find the force:

F = kx
F = (5000)(0.2)
F = 1000 N

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3 years ago

A 0.500 kg rock is whirled in a vertical circle of a radius 0.60 m . the velocity of the rock at the bottom of the swing is 4.0

OLEGan [10]

Explanation:

Centripetal acceleration is:

a = v² / r

a = (4.0 m/s)² / 0.60 m

a = 26.6 m/s²

7 0

4 years ago

A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1

bija089 [108]

1.

Answer:

Part a)

$\rho = 1.35 \times 10^{-5}$

Part b)

$\alpha = 1.12 \times 10^{-3}$

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

$i = 18.7 A$

V = 17.0 volts

now we will have

$V = I R$

$17.0 = 18.7 R$

R = 0.91 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho = 1.35 \times 10^{-5}$

Part b)

Now at higher temperature we have

$V = I R$

$17.0 = 17.3 R$

R = 0.98 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho' = 1.46 \times 10^{-5}$

so we will have

$\rho' = \rho(1 + \alpha \Delta T)$

$1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))$

$\alpha = 1.12 \times 10^{-3}$

2.

Answer:

Part a)

$i = 1.55 A$

Part b)

$v_d = 1.4 \times 10^{-4} m/s$

Explanation:

Part a)

As we know that current density is defined as

$j = \frac{i}{A}$

now we have

$i = jA$

Now we have

$j = 1.90 \times 10^6 A/m^2$

$A = \pi(\frac{1.02 \times 10^{-3}}{2})^2$

so we will have

$i = 1.55 A$

Part b)

now we have

$j = nev_d$

so we have

$n = 8.5 \times 10^{28}$

$e = 1.6 \times 10^{-19} C$

so we have

$1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d$

$v_d = 1.4 \times 10^{-4} m/s$

8 0

4 years ago

Which statements describe processes involved in the accretion of planetesimals and protoplanets? Check all that apply.

MakcuM [25]

Answer:

The correct options are;

Both involve the formation of solid particles from nebular materials

Both involve the work of gravitational push on nebular materials

Explanation:

Planetesimals are thought to be the product of grains of cosmic dusts that are found in the debris and protoplanetary discs, such that hundreds of planet forming embrayos are considered to be the result of the collisions of planetesimals that collide with each other to form larger embrayos

Protoplanets is a large planetary body with a stratified interior due to internal melting that has taken place. They originate in the protoplanetary discs from the collision of planetesimals that are up to a kilometer in size.

4 0

3 years ago

Read 2 more answers

Refrigerant 134a enters a well-insulated nozzle at 200 lbf/in.2, 200°F, with a velocity of 120 ft/s and exits at 50 lbf/in.2 wit

riadik2000 [5.3K]

Answer:

x = 0.75801 = 75.801%

T_2 = 72..78 degree F

Explanation:

From superheated R 134 a properties table

At 200 lb/in^2 and 200 degree F

$h_1 = 138.99 Btu/lbm$

steady flow energy equation is givena s

$h_1 + \frac{v_1^2}{2} = h_2 + \frac{v_2^2}{2}$

$138.99 + \frac{120^2}{2\times 25037} = h_2 + \frac{1500^2}{2 \times 25037}$

$h_2 = 94.344 Btu/lbm$

At 90 lb/in2 Tsat = 72.78 degree F

$h_f = 35.715 Btu/lbm$

hfg = 77.345 Btu/lbm

h = hf + x hfg

$94.344 = 35.715+ x \times 77.345$

solving for x we get

x = 0.75801 = 75.801%

$T_2 = 72..78 degree F$

5 0

3 years ago