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4 years ago

7

Check Your UnderstandingSuppose the radius of the loop-the-loop inExample 7.9is 15 cm and thetoy car starts from rest at a heigh

t of 45 cm above the bottom. What is its speed at the top of the loop

1 answer:

Pachacha [2.7K]4 years ago

5 0

Answer:

v = 1.7 m/s

Explanation:

By applying conservation of energy principle in this situation, we know that:

Loss in Potential Energy of Car = Gain in Kinetic Energy of Car

mgΔh = (1/2)mv²

2gΔh = v²

v = √(2gΔh)

where,

v = velocity of car at top of the loop = ?

g = 9.8 m/s²

Δh = change in height = 45 cm - Diameter of Loop

Δh = 45 cm - 30 cm = 15 cm = 0.15 m

Therefore,

v = √(2)(9.8 m/s²)(0.15 m)

<u>v = 1.7 m/s</u>

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You have 10 ohm and a 100 ohm resistor in parallel. You place this equivalent resistance in series with an LED, which is rated t

Nataly [62]

Answer:

Approximately $\rm 2.0\; V$ .

Approximately $\rm 30 \; mA$ . (assumption: the LED here is an Ohmic resistor.)

Explanation:

The two resistors here $R_1= 10\; \Omega$ and $R_2= 100\; \Omega$ are connected in parallel. Their effective resistance would be equal to

$\displaystyle \frac{1}{\dfrac{1}{R_1} + \dfrac{1}{R_2}} = \frac{1}{\dfrac{1}{10} + \dfrac{1}{100}} = \frac{10}{11} \; \Omega$ .

The current in a serial circuit is supposed to be the same everywhere. In this case, the current through the LED should be $20\; \rm mA = 0.020\; \rm A$ . That should also be the current through the effective $\displaystyle \rm \frac{10}{11} \; \Omega$ resistor. Make sure all values are in standard units. The voltage drop across that resistor would be

$V = I \cdot R = 0.020 \times \dfrac{10}{11} \approx 0.182\; \rm V$ .

The voltage drop across the entire circuit would equal to

the voltage drop across the resistors, plus
the voltage drop across the LED.

In this case, that value would be equal to $1.83 + 0.182 \approx 2.0\; \rm V$ . That's the voltage that needs to be supplied to the circuit to achieve a current of $20\; \rm mA$ through the LED.

Assuming that the LED is an Ohmic resistor. In other words, assume that its resistance is the same for all currents. Calculate its resistance:

$\displaystyle R(\text{LED}) = \frac{V(\text{LED})}{I(\text{LED})}= \frac{1.83}{0.020} \approx 91.5\; \Omega$ .

The resistance of a serial circuit is equal to the resistance of its parts. In this case,

$\displaystyle R = R(\text{LED}) + R(\text{Resistors}) = 91.5 + \frac{10}{11} \approx 100\; \Omega$ .

Again, the current in a serial circuit is the same in all appliances.

$\displaystyle I = \frac{V}{R} = \frac{3}{100} \approx 0.030\; \rm A = 30\; mA$ .

7 0

3 years ago

A team of astronauts is on a mission to land on and explore a large asteroid. In addition to collecting samples and performing e

GenaCL600 [577]

Answer:

463.4 m/s

Explanation:

The escape velocity on the surface of a planet/asteroid is given by

$v=\sqrt{\frac{2GM}{R}}$ (1)

where

G is the gravitational constant

M is the mass of the planet/asteroid

R is the radius of the planet/asteroid

For the asteroid in this problem, we know

$\rho=4.49\cdot 10^6 g/m^3$ is the density

$V=3.32\cdot 10^{12} m^3$ is the volume

So we can find its mass:

$M=\frac{\rho}{V}=(4.49\cdot 10^6 g/m^3)(3.32\cdot 10^{12}m^3)=1.49\cdot 10^{19} kg$

Also, the asteroid is approximately spherical, so its volume is given by

$V=\frac{4}{3}\pi R^3$

where R is the radius. Solving the formula for R, we find its radius:

$R=\sqrt[3]{\frac{3V}{4\pi}}=\sqrt[3]{\frac{3(3.32\cdot 10^{12}m^3)}{4\pi}}=9256 m$

So now we can use eq.(1) to find the escape velocity:

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(1.49\cdot 10^{19}kg)}{9256 m}}=463.4 m/s$

3 0

3 years ago

Velocity is the slope of the acceleration vs. time graph.<br> A.) True<br> B.)False<br> (apex)

Mekhanik [1.2K]

Velocity vs. time graph shows the acceleration as a slope whereas displacement vs. time graph shows the velocity as a slope. So, the given statement is false.

Answer: Option B

<u>Explanation:</u>

To understand the acceleration graphically, consider the x axis of the graph as the run and the y axis as the velocity rise. Now, as we all know that,

$\text {Acceleration}=\frac{\text {Change in velocity }}{\text {Time interval}}$

We can estimate this through the graph. let's draw the motion of an object with time if it's velocity is changing in every second by 4 m/s. Now if we draw this on graph, we will see that there is a slope between the two corresponding values of time and velocity. This slope defines the acceleration for the object with time.

Now, in the same way, if we draw a distance and time graph respective to the y and x axis; we'll get a slope which defines the velocity of the object i.e. change in distance with time.

Hence, with a velocity vs time graph, we get the slope for acceleration whereas with the distance and time graph, we get the slope for velocity. So both the cases, we see there is no velocity slope on an acceleration and time graph. Hence the statement is false.

8 0

3 years ago

Read 2 more answers

The small spherical planet called "Glob" has a mass of 7.88×10^18 kg and a radius of 6.32×10^4 m. An astronaut on the surface of

Oksi-84 [34.3K]

Answer: The small spherical planet called "Glob" has a mass of 7.88×1018 kg and a radius of 6.32×104 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×103 m, above the surface of the planet, before it falls back down.

1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.

2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.

Explanation: To find the answer, we need to know about the different equations of planetary motion.

<h3>How to find the initial speed of the rock as it left the astronaut's hand?</h3>

We have the expression for the initial velocity as,

$v=\sqrt{2gh}$

Thus, to find v, we have to find the acceleration due to gravity of glob. For this, we have,

$g_g=\frac{GM}{r^2} =\frac{6.67*10^{-11}*7.88*10^{18}}{(6.32*10^4)^2}= 0.132$

Now, the velocity will become,

$v=\sqrt{2*0.132*1.44*10^3} =19.46 m/s$

<h3>How to find the speed of the satellite?</h3>

As we know that, by equating both centripetal force and the gravitational force, we get the equation of speed of a satellite as,

$v=\sqrt{\frac{GM}{r} } =\sqrt{\frac{6.67*10^{-11}*7.88*10^{18}}{1.45*10^5} } =3.624km/s$

Thus, we can conclude that,

1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.

2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.

Learn more about the equations of planetary motion here:

brainly.com/question/28108487

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3 0

2 years ago

Read 2 more answers

A 65-kg bungee jumper, who is attached to one end of an 85-m long bungee cord that has its other end tied to a bridge, jumps off

irinina [24]

Answer:

The impulse delivered to the bungee jumper is 1.32 kN.s

Explanation:

The situation can be shown graphically as shown in the figure.

Impulse delivered to the bungee jumper = Area under the curve.

The curve represents a triangle and the area of traiangle = (1/2)base×height

The base of the triangle from the graph = 1.2 seconds.

The height of the triangle from the graph = 2.2 kN

Thus,

<u>Impulse = (1/2)×(1.2 seconds)×(2.2 kN) = 1.32 kN.s</u>

3 0

3 years ago