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2 years ago

50 points and brainliest!!!! plz help me

Physics

2 answers:

antoniya [11.8K]2 years ago

3 0

decrease yeah free 100 percent

disa [49]2 years ago

3 0

Answer:

The amount of gas molecules in the air decreaces while traveling up through the troposphere—the air becomes less dense than air nearer to sea level.

Explanation:

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A 13-kg sled is moving at a speed of 3.0 m/s. At which of the following speeds will the sled have twice as much kinetic energy?

AysviL [449]

K.E. = 1/2 mv²
K.E. is directly proportional to v^2
So, when K.E. increase by 2, K.E. increase by root. 2
v' = 1.41v
original v value was 3 so, final would be:
v' = 1.41*3 = 4.23
After round-off to it's tenth value, it will be:
v' = 4.2

So, option B is your answer!

Hope this helps!

7 0

3 years ago

An initially motionless test car is accelerated to 115 km/h in 8.58 s before striking a simulated deer. The car is in contact wi

hoa [83]

Answer:

a) a = 3.72 m / s², b) a = -18.75 m / s²

Explanation:

a) Let's use kinematics to find the acceleration before the collision

v = v₀ + at

as part of rest the v₀ = 0

a = v / t

Let's reduce the magnitudes to the SI system

v = 115 km / h (1000 m / 1km) (1h / 3600s)

v = 31.94 m / s

v₂ = 60 km / h = 16.66 m / s

et's calculate

a = 31.94 / 8.58

a = 3.72 m / s²

b) For the operational average during the collision let's use the relationship between momentum and momentum

I = Δp

F Δt = m v_f - m v₀

F = $\frac{m ( v_f - v_o)}{t}$

F = m [16.66 - 31.94] / 0.815

F = m (-18.75)

Having the force let's use Newton's second law

F = m a

-18.75 m = m a

a = -18.75 m / s²

4 0

2 years ago

A person is riding a bicycle, and its wheels have an angular velocity of 10.7 rad/s. Then, the brakes are applied and the bike i

Scilla [17]

Answer:

(a) t = 22.9 s

(b) α= - 0.467 rad/s²

Explanation:

The uniformly accelerated circular movement, is a circular path movement in which the angular acceleration is constant.

We apply the equations of circular motion uniformly accelerated :

ωf ²= ω₀² + 2*α*θ Formula (1)

ωf= ω₀ + α*t Formula (2)

Where:

θ : angle that the body has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

t : time interval (s)

ω₀ : initial angular speed ( rad/s)

ωf : final angular speed ( rad/s)

Data

θ = 19.5 revolutions : angular displacement of each wheel or angle that the wheel has rotated in a given time interval

ω₀= 10.7 rad/s : initial angular speed of the Wheel ( rad/s)

ωf = 0 : final angular speed of the Whee( rad/s)

Calculating of the angular acceleration (α )

We replace data in the fómula (1),considering that 1 revolution is equal to 2π radians :

ωf ²= ω₀² + 2*α*θ

(0 )²= (10.7)² + 2*α*(19.5*2*π )

0= 114.49 + (245.04)*α

-114.49 = (245.04)*α

α= (-114.49) /(245.04)

α= -114.49 /(245.04)

α= -0.467 rad/s²

Time does it take for the bike to come to rest

We replace data in the formula (2)

ωf = ω₀ + α*t

0 = 10.7 + -0.467*t

-10.7 = - 0.467*t we multiply by (-1) both sides of the equation :

10.7 = 0.467*t

t = 10.7 / 0.467

t = 22.9 s

3 0

3 years ago

A 17926-lb truck enters an emergency exit ramp at a speed of 75.6 ft/s. It travels for 6.4 s before its speed is reduced to 30.3

I am Lyosha [343]

Answer:

$F_{braking}=337299 pdl$

Explanation:

Impulse-Momentum relation:

$I=\Delta p\\ F_{total}*t=m(v_{f}-v{o})$

$F_{total}=-F_{braking}+mgsin{\theta}$

We solve the equations in order to find the braking force:

$F_{braking}=m(v_{o}-v{f})/t+mgsin{\theta}=17926(75.6-30.3)/6.4+17926*32.17*sin21.4=337299 pdl$

6 0

3 years ago

Can i get help for the parts for the two questions? MATHPHYSSSS

Anestetic [448]

Explanation:

003 (part 1 of 2)

Pressure is force divided by area.

P = F / A

P = (117 kg × 9.8 m/s²) / (2 × (0.05 m)²)

P = 229,320 Pa

003 (part 2 of 2)

There are approximately 6895 Pa in 1 psi.

P = 229,320 Pa × (1 psi / 6895 Pa)

P = 33.3 psi

004 (part 1 of 2)

Since the collisions are elastic, the angle of reflection is the same as the angle of incidence (it bounces off at the same angle).

Impulse = change in momentum

F Δt = m Δv

F (36 s) = (300 × 0.003 kg) (5.2 sin 57° m/s − (-5.2 sin 57° m/s))

F = 0.218 N

004 (part 2 of 2)

Pressure is force over area.

P = F / A

P = 0.218 N / 0.712 m²

P = 0.306 N/m²

7 0

3 years ago