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oee [108]
2 years ago
15

At a drag race, a jet car travels 1/4 mile in 5.2 seconds. What is the final speed of the

Physics
1 answer:
Sedbober [7]2 years ago
3 0

Answer:

1609.3

Explanation:

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A helicopter travels west at 80 mph. It is moving above a car traveling on a highway at 80 mph. Given this information, you can
gavmur [86]

Answer:

d. at the same velocity

Explanation:

I will assume the car is also travelling westward because it was stated that the helicopter was moving above the car. In that case, it depends where the observer is. If the observer is in the car, the helicopter would look like it is standing still ( because both objects have the same velocity). If the observer is on the side of the road, both objects would be travelling at the same velocity. Also recall that, velocity is a vector quantity; it is direction-aware. Velocity is the rate at which the position changes but speed is the rate at which object covers distance and it is not direction wise. Hence velocity is the best option.

5 0
1 year ago
A 5-ft-tall person walks away from the wall at a rate of 2 ft/sec. A spotlight is located on the ground 40 ft from the wall. How
AveGali [126]

Answer:

The rate of change of the height is - 4 ft/s

Solution:

As per the question:

Height of the person, y = 5 ft

The rate at which the person walks away, \frac{dx}{dt} = 4\ ft/s

Distance of the spotlight from the wall, x = 40 ft

Now,

To calculate the rate of change in the height, \frac{dy}{dt} of the person when, x = 10 m:

From fig 1.

\Delta ABC[\tex] ≈ [tex]\Delta PQC[\tex]Thus[tex]\frac{BC}{AB} = \frac{PQ}{QC}

\frac{y}{40} = \frac{5}{x}

xy = 200                                                                       (1)

Differentiating the above eqn w.r.t time t:

x\frac{dy}{dt} + y\frac{dx}{dt} = 0

Thus

\frac{dy}{dt} = - \frac{y}{x}\frac{dx}{dt}              (2)

From eqn (1):

When x = 10 ft

10y = 200

y = 20 ft

Using eqn (2):

\frac{dy}{dt} = - \frac{20}{10}\times 2 = - 4\ ft

8 0
2 years ago
No official standard exists for extinguisher-cylinder colors. that said, a red cylinder is typically associated with which type
Lorico [155]
Fire extinguishers are typically red. Goes along with the red fire trucks and red being a color of emergency. While there isn't any standard, if we decided to change the colors of the fire extinguisher, people would be very confused and given the context, it could cost people's lives, so I'm thinking red will be the color of fire extinguishers for a while to come.
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8 0
2 years ago
Suppose that we use a heater to boil liquid nitrogen (N2 molecules). 4480 J of heat turns 20 g of liquid nitrogen into gas. Note
love history [14]

Answer:

The energy is 9.4\times10^{-21}\ J

(a) is correct option

Explanation:

Given that,

Energy = 4480 j

Weight of nitrogen = 20 g

Boil temperature = 77 K

Pressure = 1 atm

We need to calculate the internal energy

Using first law of thermodynamics

Q=\Delta U+W

Q=\Delta U+nRT

Put the value into the formula

4480=\Delta U+\dfrac{20}{28}\times8.314\times77

\Delta U=4480-\dfrac{20}{28}\times8.314\times77

\Delta U=4022.73\ J

We need to calculate the number of molecules in 20 g N₂

Using formula of number of molecules

N=n\times \text{Avogadro number}

Put the value into the formula

N=\dfrac{20}{28}\times6.02\times10^{23}

N=4.3\times10^{23}

We need to calculate the energy

Using formula of energy

E=\dfrac{\Delta U}{N}

Put the value into the formula

E=\dfrac{4022.73}{4.3\times10^{23}}

E=9.4\times10^{-21}\ J

Hence, The energy is 9.4\times10^{-21}\ J

4 0
1 year ago
A bulldozer attempts to drag a log weighing 500 N along the rough horizontal ground. The cable attached to the log makes an angl
Gemiola [76]

Answer:

T= 224.01 N

Explanation:

in imminent motion we have to :

  • The frictional force reaches its maximum value
  • The system is in balance of forces

Data

W=  500 N :  weight of the log

μs = 0.5

μk = 0.35

α = 30°above the ground :  angle of the cable attached to the log

Newton's first law to the log:

∑F =0 Formula (1)

∑F : algebraic sum of the forces in Newton (N)

Forces acting on the log

T: cable tension for impending movement

N: normal force

W : weight

f: frictional force , f= μsN

We apply the formula (1)

∑Fx=0

Tx-f = 0

Tcosα-μsN=0

Tcos30°-0.5N=0 Equation (1)

∑Fy=0

N+Ty-W=0

N+Tsin30°-500=0

N= 500-Tsin30°  Equation (2)

We replace the value of N of the Equation  (2) in the equation (1)

Tcos30°-0.5(500-Tsin30°) = 0

Tcos30°+0.5Tsin30° = 0.5*500

T( cos30°+0.5*sin30°) = 250

(1.116) T = 250

T= 250/1.116

T= 224.01 N

6 0
1 year ago
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