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3 years ago

Please need help with this

Physics

1 answer:

NeTakaya3 years ago

7 0

Note that:

The gravitational potential energy = $mgh$

where m: is the mass, g: the acceleration due to the gravity and h is the height from the earth surface

Then, we can increase the gravitational potential energy by increasing the mass or the height from the earth surface

In our question, we can increase the gravitational potential energy by

A) Strap a boulder to the car so that it wights more.

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We’re measuring the velocity, so that we can find the change in kinetic energy. Where should you place the photogate to measure

7nadin3 [17]

Answer:

minimum two photogate

Explanation:

The photogate allows us to activate the measurement of time, so we must place at least two of them, one at the beginning of the movement to measure the initial time and another at the end of the movement to measure the final time, of course we must have measured the distance. This gives us the average speed between the two photocells

For measurements with more changes curves, slopes, one photogate must be placed at the beginning of the curve or slope and another at the end.

4 0

4 years ago

A circuit contains an EMF source, a resistor R, a capacitor C, and an open switch in series. The capacitor initially carries zer

Flura [38]

Answer:

t = 1.098*RC

Explanation:

In order to calculate the time that the capacitor takes to reach 2/3 of its maximum charge, you use the following formula for the charge of the capacitor:

$Q=Q_{max}[1-e^{-\frac{t}{RC}}]$ (1)

Qmax: maximum charge capacity of the capacitor

t: time

R: resistor of the circuit

C: capacitance of the circuit

When the capacitor has 2/3 of its maximum charge, you have that

Q=(2/3)Qmax

You replace the previous expression for Q in the equation (1), and use properties of logarithms to solve for t:

$Q=\frac{2}{3}Q_{max}=Q_{max}[1-e^{-\frac{t}{RC}}]\\\\\frac{2}{3}=1-e^{-\frac{t}{RC}}\\\\e^{-\frac{t}{RC}}=\frac{1}{3}\\\\-\frac{t}{RC}=ln(\frac{1}{3})\\\\t=-RCln(\frac{1}{3})=1.098RC$

The charge in the capacitor reaches 2/3 of its maximum charge in a time equal to 1.098RC

6 0

4 years ago

the speed of a train is decreased in a uniform rate from 96 km/h to 48km/h through a distance of 800m. calculate the distance co

GREYUIT [131]

Answer:

1066.67 m

Explanation:

Given:

v₀ = 96 km/h = 26.67 m/s

v = 48 km/h = 13.33 m/s

Δx = 800 m

Find: a

v² = v₀² + 2aΔx

(13.33 m/s)² = (26.67 m/s)² + 2a (800 m)

a = -0.333 m/s²

Given:

v₀ = 26.67 m/s

v = 0 m/s

a = -0.333 m/s²

Find: Δx

v² = v₀² + 2aΔx

(0 m/s)² = (26.67 m/s)² + 2 (-0.333 m/s²) Δx

Δx = 1066.67 m

Round as needed.

7 0

3 years ago

What is the base number for the metric system of units? In other words, by what number do we multiply to move up in scale ? ____

Andru [333]

Answer:

Explanation:

Each time you want to move up the scale, the number is multiplied by 10. The opposite is true when moving down; you must divide by 10.

6 0

3 years ago

Read 2 more answers

How long does a car (1000 kg) have a speed of 30 m/s from a rest if the engine power is 10kw

Lemur [1.5K]

Answer: 90.1 s

Explanation:

Use equation for power:

P=F*V

Use eqation for force:

F=ma

F---force

V---velocity

Vr=om/s

V=30m/s

m=1000kg

P=10000W

---------------------------

P=FV

F=P/V

F=10000W/30m/s

F=333.33N

Use equation for force to find accelartaion.

F=ma

a=F/m

a=333.33N/1000kg

a=0.333 m/s²

Use equation for accelaration to find out time:

a=(V-Vs)/t

t=(V-Vs)/a

t=(30m/s)/(0.333m/s²)

t=90.09 s≈90.1 s

------------------------

5 0

3 years ago