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Luda [366]
3 years ago
13

Can someone help me with this real quick?

Physics
1 answer:
ch4aika [34]3 years ago
4 0

Answer:

7,546 J

Explanation:

recall that Potential energy is given by

P.E = mgΔh

where m = 70kg (given)

g = 9.8 m/s² (acceleration due to gravity)

Δh = change in height

= distance from top of building to top of car

= height of building - height of car

= (5+8) - 2

= 11m

substituting all these into the equation:

P.E = mgΔh

= 70 x 9.8 x 11

= 7,546 J

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A car traveling 90 km/hr is 100 m behind a truck traveling 50 km/hr. How long will it take the car to reach the truck?
Ber [7]

v = speed of car = 90 km/h

u = speed of truck = 50 km/h

d = initial separation distance = 100 m = 0.1 km

They meet at time t such that

vt = d + ut

t(v - u) = d

t = d/(v - u) = (0.1 km) / [(90 - 50) km/h] = 0.0025 hours

8 0
4 years ago
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A ball player catches a ball 3.24 s after throwing it vertically upward.With what speed did he throw it?
Firdavs [7]
Doesn't seem like we know much here, but we can answer it. Let's talk about what we know. We know it takes 3.24 s for the ball to go up and drop back down again. We know that gravity is the only force acting after the ball leaves the hand, so a = 9.8 m/s^2 (we'll say it's negative in our equations because down being negative is intuitive). We also know that it stops moving for a brief moment at the top of the arc, where v = 0 m/s. Because gravity is the only force, and it slows it down on the way up at the same rate it speeds it up on the way down and the distance covered in upward and downward motion is the same, we can confidently say that it will reach the top of its arc (where v = 0 and it turns around) in half the total time it is in the air, so it takes 1.62 s to reach the peak. Now we can use a kinematics equation, let's use vf = vi + a*t, where vf is final velocity and is 0, vi is initial velocity and is some unknown v we need to solve for, a is acceleration and is -9.8 m/s^2 and t is time and since this is just to the top of the arc, we'll use half the time so 1.62 s. We can solve for vi and plug stuff in like so: v = -a*t = -(-9.8m/s^2)*(1.62s) = 15.876 m/s.
8 0
4 years ago
As an aid in working this problem, consult Concept Simulation 11.1. The volume flow rate in an artery supplying the brain is 3.5
klemol [59]

Answer

given,

flow from the artery = 3.5 x 10⁻⁶ m³/s

Radius of artery = 5.80 x 10⁻³m

area = π R²

       = π x (5.8 x 10⁻³)²

       = 1.06 x 10⁻⁴ m²

v = \dfrac{Q}{A}

v = \dfrac{3.5 \times 10^{-6}}{1.06 \times 10^{-4}}

v = 0.033 m/s

b) new velocity of flow

Radius = R' = R/4

 A V = A' V'

 R² V = R'² V'

 R^2 V = (\dfrac{R}{4})^2 V'

 V' = 16 V

 V' = 16 x 0.033

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5 0
3 years ago
An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 2.00 mm. If a
ValentinkaMS [17]

Answer:

a. 11.5kv/m

b.102nC/m^2

c.3.363pF

d. 77.3pC

Explanation:

Data given

area=7.60cm^{2}\\ distance,d=2mm\\voltage,v=23v

to calculate the electric field, we use the equation below

V=Ed

where v=voltage, d= distance and E=electric field.

Hence we have

E=v/d\\E=\frac{23}{2*10^{-3}} \\E=11.5*10^{3} v/m\\E=11.5Kv/m

b.the expression for the charge density is expressed as

σ=ξE

where ξ is the permitivity of air with a value of 8.85*10^-12C^2/N.m^2

If we insert the values we have

8.85*10^{-12} *11500\\1.02*10^{-7}C/m^{2}  \\102nC/m^{2}

c.

from the expression for the capacitance

C=eA/d

if we substitute values we arrive at

C=\frac{8.85*10^{-12}*7.6*10^{-4}}{2*10^{-3} } \\C=\frac{6.726*10^{-15} }{2*10^{-3} } \\C=3.363*10^{-12}F\\C=3.363pF

d. To calculate the charge on each plate, we use the formula below

Q=CV\\Q=23*3.363*10^{-12}\\ Q=7.73*10^{-12}\\ Q=77.3pC

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