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3 years ago

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A toy car was used in an investigation about unbalanced forces. During which of the following scenarios was the car not experien

cing an unbalanced force? *
20 points
1.The toy car speeds up as it rolls down a ramp.
2.The toy car travels forward at a constant speed.
3.The toy car slows down as it rolls up a hill.
4.The toy car changes direction at a constant speed.

2 answers:

lozanna [386]3 years ago

7 0

Number 2, the question asked which car was not experiencing a unbalanced force. number 2 says the car is traveling at a constant speed signifying that nothing was in the cars way allowing it to travel with ease

Firlakuza [10]3 years ago

4 0

Answer:

Number 2

it is using Newton’s first law.

also known as the law of inertia

look that last word up

Explanation:

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Please, someone help me I'm begging yall 1. it's estimated that 1 kg of body fat will provide 3.8 * 10^7 J of energy. A 67 kg mo

Leto [7]

Answer:

0.24 kg used up

Explanation:

He has a mass of 67 kg

The gravitational constant is 9.81 m/s^2

The distance upward is 3500 m

W = m*g*h

W = 67 * 9.81 * 3500

Work = 2,300,445 Joules

Work = 2300 kj

work = 2.30 * 10^6 joules in scientific notation.

Part B

He needs 4 times this amount to climb the mountain because the body is only 25% efficient in converting energy.

4*2.30 * 10^6 = 9.20 * 10^6 Joules of energy are therefore required.

The total amount in a kg of fat = 3.8 * 10^7 joules

x kg of fat is needed to provide 9.20.*10^6 joules

1 kg / (3.8 * 10^7 J ) = x kg / (9.20 * 10^6 J)

9.20 * 10 ^6 * 1 = 3.8 * 10^7 *x

9.20 * 10 ^6 / 3.8 * 10^7 = x

x = 0.24 kg of fat are needed

7 0

3 years ago

If acceleration is zero what statement about velocity is true *

Wittaler [7]

Answer:

Option"B" is correct.

Explanation:

when a body move with constant velocity then acceleration is zero.

6 0

3 years ago

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Un objeto tiene una masa de 120 kg y un volumen de 5 m3

melisa1 [442]

Answer: dont ever worry mate

Explanation:

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3 years ago

The moon is 3x10^5 km away from Nepal and the mass of the moon is 7x10^22 kg. Calculate the force with which the Moon pulls ever

hammer [34]

Answer:

Approximately $5.19 \times 10^{-5}\; \rm N$ .

Explanation:

Let $G$ denote the gravitational constant. ( $G \approx 6.67 \times 10^{-11} \; \rm N \cdot kg^{-2} \cdot m^{2}$ .)

Let $M$ and $m$ denote the mass of two objects separated by $r$ .

By Newton's Law of Universal Gravitation, the gravitational attraction between these two objects would measure:

$\displaystyle F = \frac{G \cdot M \cdot m}{r^{2}}$ .

In this question: $M = 7 \times 10^{22}\; \rm kg$ is the mass of the moon, while $m = 1\; \rm kg$ is the mass of the water. The two are $r = 3\times 10^{5}\; \rm km$ apart from one another.

Important: convert the unit of $r$ to standard units (meters, not kilometers) to reflect the unit of the gravitational constant $G$ .

$\displaystyle r = 3 \times 10^{5}\; \rm km \times \frac{10^{3}\; \rm m}{1\; \rm km} = 3 \times 10^{8}\; \rm m$ .

$\begin{aligned} F &= \frac{G \cdot M \cdot m}{r^{2}} \\ &= \frac{6.67 \times 10^{-11}\; \rm N \cdot kg^{-2}\cdot m^{2} \times 7 \times 10^{22}\; \rm kg \times 1\; \rm kg}{(3 \times 10^{8}\; \rm m)^{2}} \\ &\approx 5.19 \times 10^{-5} \; \rm N\end{aligned}$ .

5 0

3 years ago

Effciency of a lever is never 100% or more. why?Give reason

Troyanec [42]

Answer:

Ideally, the work output of a lever should match the work input. However, because of resistance, the output power is nearly always be less than the input power. As a result, the efficiency would go below $100\%$ .

Explanation:

In an ideal lever, the size of the input and output are inversely proportional to the distances between these two forces and the fulcrum. Let $D_\text{in}$ and $D_\text{out}$ denote these two distances, and let $F_\text{in}$ and $F_\text{out}$ denote the input and the output forces. If the lever is indeed idea, then:

$F_\text{in} \cdot D_\text{in} = F_\text{out} \cdot D_\text{out}$ .

Rearrange to obtain:

$\displaystyle F_\text{in} = F_\text{out} \cdot \frac{D_\text{out}}{D_\text{in}}$

Class two levers are levers where the perpendicular distance between the fulcrum and the input is greater than that between the fulcrum and the output. For this ideal lever, that means $D_\text{in} > D_\text{out}$ , such that $F_\text{in} < F_\text{out}$ .

Despite $F_\text{in} < F_\text{out}$ , the amount of work required will stay the same. Let $s_\text{out}$ denote the required linear displacement for the output force. At a distance of $D_\text{out}$ from the fulcrum, the angular displacement of the output force would be $\displaystyle \frac{s_\text{out}}{D_\text{out}}$ . Let $s_\text{in}$ denote the corresponding linear displacement required for the input force. Similarly, the angular displacement of the input force would be $\displaystyle \frac{s_\text{in}}{D_\text{in}}$ . Because both the input and the output are on the same lever, their angular displacement should be the same:

$\displaystyle \frac{s_\text{in}}{D_\text{in}} =\frac{s_\text{out}}{D_\text{out}}$ .

Rearrange to obtain:

$\displaystyle s_\text{in}=s_\text{out} \cdot \frac{D_\text{in}}{D_\text{out}}$ .

While increasing $D_\text{in}$ reduce the size of the input force $F_\text{in}$ , doing so would also increase the linear distance of the input force $s_\text{in}$ . In other words, $F_\text{in}$ will have to move across a longer linear distance in order to move $F_\text{out}$ by the same $s_\text{out}$ .

The amount of work required depends on both the size of the force and the distance traveled. Let $W_\text{in}$ and $W_\text{out}$ denote the input and output work. For this ideal lever:

$\begin{aligned}W_\text{in} &= F_\text{in} \cdot s_\text{in} \\ &= \left(F_\text{out} \cdot \frac{D_\text{out}}{D_\text{in}}\right) \cdot \left(s_\text{out} \cdot \frac{D_\text{in}}{D_\text{out}}\right) \\ &= F_\text{out} \cdot s_\text{out} = W_\text{out}\end{aligned}$ .

In other words, the work input of the ideal lever is equal to the work output.

The efficiency of a machine can be measured as the percentage of work input that is converted to useful output. For this ideal lever, that ratio would be $100\%$ - not anything higher than that.

On the other hand, non-ideal levers take in more work than they give out. The reason is that because of resistance, $F_\text{in}$ would be larger than ideal:

$\displaystyle F_\text{in} = F_\text{out} \cdot \frac{D_\text{out}}{D_\text{in}} + F(\text{resistance})$ .

As a result, in real (i.e., non-ideal) levers, the work input will exceed the useful work output. The efficiency will go below $100\%$ ,

4 0

3 years ago