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lyudmila [28]
3 years ago
5

Best Answer gets Brainiest

Physics
1 answer:
Alexeev081 [22]3 years ago
6 0

Answer:

yoo yoo

24

Explanation:

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A car drives 100 km north, 50 km east, then 10 km south. What is the car's displavement?​
Over [174]

Answer:

\sqrt{ ({40}^{2} }  +  {50}^{2} ) =  \sqrt{(1600 + 2500)}  =  \sqrt{4100}  = 64.03

6 0
3 years ago
A car travels across Texas m miles at the rate of t miles per hour. How many hours does the trip take??
Marianna [84]

Answer: The trip takes \frac{m}{t}hours

Explanation:

Velocity V is the variation of the position of a body (distance traveled d) with time T:

V=\frac{d}{T}

In this case, the car travels a distance d=m miles at a velocity V=t \frac{miles}{hour} and we need to find the time it takes the trip.

Isolating  T:

T=\frac{d}{V}=\frac{m miles}{t \frac{miles}{hour}}

Finally:

T=\frac{m}{t}hours

8 0
3 years ago
Zookeepers carry a stretcher that holds a sleeping lion. The total mass of the stretcher and lion is 175 kg. The lion's forward
ASHA 777 [7]

Answer:

350 N

Explanation:

Newton's second law:

∑F = ma

∑F = (175 kg) (2 m/s²)

∑F = 350 N

6 0
3 years ago
Which is a sub-atomic particle?
Natalka [10]
A particle that is smaller than an atom or a cluster of particles.
4 0
3 years ago
A mass weighing 14 pounds stretches a spring 2 feet. The mass is attached to a dashpot device that offers a damping force numeri
Elodia [21]

Answer:

The motion is over-damped when λ^2 - w^2 > 0 or when b^{2} > 0.86

The motion is critically when λ^2 - w^2 = 0 or when b^{2} = 0.86

The motion is under-damped when λ^2 - w^2 < 0 or when b^{2} < 0.86

Explanation:

Using the newton second law

k is the spring constante

b positive damping constant

m mass attached

m\frac{d^{2} x}{dt^{2}} = - kx - b\frac{dx}{dt}

x(t) is the displacement from the equilibrium position

\frac{d^{2} x}{dt^{2}} +\frac{b}{m}\frac{dx}{dt} + \frac{k}{m}x = 0

Converting units of weights in units of mass (equation of motion)

m = \frac{W}{g} = \frac{14}{32} = 0.43 slug

From hook's law we can calculate the spring constant k

k = \frac{W}{s} = \frac{14}{2} = 7 lb/ft

If we put m and k into the DE, we get

\frac{d^{2} x}{dt^{2}} +\frac{b}{0.43}\frac{dx}{dt} + 16.28x = 0

Denoting the constants

2λ = \frac{b}{m} = \frac{b}{0.43}

λ = b/0.215

w^{2} = \frac{k}{m} = 16.28

λ^2 - w^2 = \frac{b^{2} }{0.046} - 16.28

This way,

The motion is over-damped when λ^2 - w^2 > 0 or when b^{2} > 0.86

The motion is critically when λ^2 - w^2 = 0 or when b^{2} = 0.86

The motion is under-damped when λ^2 - w^2 < 0 or when b^{2} < 0.86

3 0
3 years ago
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