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2 years ago

11

a car accelerates along a straight road from rest to 75km/h in 5.0s. What is the magnitude of its average acceleration? show up

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1 answer:

kykrilka [37]2 years ago

5 0

Answer:

4.186 m/s^2

Explanation:

First, convert km/hr to m/s:

(75 km/hr)(1000m/1km)(1hr/60min)(1min/60s) = 20.83 m/s

Then, divide 20.93 m/s by 5.0s

(20.93 m/s) / (5.0s) = 4.186 m/s^2

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A 20-Kg child is on a swing attached to 3.0 m-long chains. The child swings back and forth, swinging out to a 60-degree angle. (

kvv77 [185]

Answer:

v = 29.4 m / s

Explanation:

For this exercise we can use the conservation of mechanical energy

Lowest starting point.

Em₀ = K = ½ m v²

final point. Higher

$Em_{f}$ = U = m g h

Let's use trigonometry to lock her up

cos 60 = y / L

y = L cos 60

Height is the initial length minus the length at the maximum angle

h = L - L cos 60

h = L (1- cos 60)

energy is conserved

Em₀ = Em_{f}

½ m v² = mgL (1 - cos 60)

v = 2g L (1- cos 60)

let's calculate

v² = 2 9.8 3.0 (1- cos 60)

v = 29.4 m / s

6 0

3 years ago

Lasers can be constructed that produce an extremely high intensity electromagnetic wave for a brief time—called pulsed lasers. T

alex41 [277]

Answer:

30643 J

Explanation:

$\mu_0$ = Vacuum permeability = $4\pi \times 10^{-7}\ H/m$

t = Time taken = 1 ns

c = Speed of light = $3\times 10^8\ m/s$

$E_0$ = Maximum electric field strength = $1.52\times 10^{11}\ V/m$

A = Area = $1\ mm^2$

Magnitude of magnetic field is given by

$B_0=\dfrac{E_0}{c}\\\Rightarrow B_0=\dfrac{1.52\times 10^{11}}{3\times 10^8}\\\Rightarrow B_0=506.67\ T$

Intensity is given by

$I=\dfrac{cB_0^2}{2\mu_0}\\\Rightarrow I=\dfrac{3\times 10^8\times 506.67^2}{2\times 4\pi \times 10^{-7}}\\\Rightarrow I=3.0643\times 10^{19}\ W/m^2$

Power, intensity and time have the relation

$E=IAt\\\Rightarrow E=3.0643\times 10^{19}\times 1\times 10^{-6}\times 1\times 10^{-9}\\\Rightarrow E=30643\ J$

The energy it delivers is 30643 J

4 0

3 years ago

1)the car's engine power is 44000W. Explain this number in a physical sense

Ratling [72]

Answer:

1) It expresses the rate (top speed) at which it can move with time.

2) P = 20 W

3) h = 18 km

Explanation:

1) Power is the rate of transfer of energy.

⇒ Power = $\frac{Energy(or workdone)}{Time}$

i.e P = $\frac{E}{t}$

Thus a car's engine power is 44000W implies that the engine of the car can propel the car at this rate. This expresses the rate (top speed) at which it can move with time.

2) m = 400g = 0.4 kg

t = 20 s

h = 100m

g = 10 m/ $s^{2}$

P = $\frac{mgh}{t}$

= $\frac{0.4*10*100}{20}$

= $\frac{400}{20}$

P = 20 W

3) u = 600 m/s

g = 10 m/ $s^{2}$

From the third equation of free fall,

$V^{2}$ = $U^{2}$ - 2gh

V is the final velocity, U is the initial velocity, h is the height.

0 = $(600)^{2}$ - 2 x 10 x h

0 = 360000 - 20h

20h = 360000

h = $\frac{360000}{20}$

= 18000

h = 18 km

The maximum height of the bullet would be 18 km.

3 0

3 years ago

Two objects exert a gravitational force of 3 N on one another when they are 10 m apart. What would that force be if the distance

jeka94

Gravitational force = G ( m1 m2 ) / r²
3 = G ( m1 m2 ) / ( 10 )²x = G ( m1 m2 ) / ( 5 )²We shall divide those two equations:3 / x = 1/100 / 1/25 = 25 / 100 = 1 / 4x · 1 = 3 · 4x = 12Answer:C. 12 N

8 0

3 years ago

A 75 W lightbulb is being run on a 110 V outlet. Determine the resistance of the lightbulb. Provide a detailed description of th

Sauron [17]

Power is defined as

P = I*V

where I is the current and V is the voltage

Ohm's law gives us the relation betwen Voltage and current in a resistive component

V = I*R , Then

P = V² / R

We solve for R,

R = (110 V)²/ 75W = 161.33 ohms

8 0

3 years ago