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2 years ago

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2. An 85.0 kg cart is rolling along a level road at 9.00 m/s. The cart encounters a hill and coasts up the hill. a) Assuming the

movement is frictionless, at what vertical height will the cart come to rest? b) Do you need to know the mass of the cart to solve this problem?

1 answer:

Luda [366]2 years ago

3 0

Yes please give me the mass of it.

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While traveling from Boston to Hartford, Person A drives at a constant speed of 55 mph for the entire trip. Person B drives at 6

Ne4ueva [31]

Answer:

B will take 1.034 times the time of A from Boston to Hartford.

Explanation:

Let the distance from Boston to Hartford be S.

Person A drives at a constant speed of 55 mph for the entire trip,

Time taken by person A

$t_A=\frac{S}{55}$

Person B drives at 65 mph for half the distance and then drives 45 mph for the second half of the distance.

Time taken by person B

$t_B=\frac{\frac{S}{2}}{65}+\frac{\frac{S}{2}}{45}=\frac{S}{130}+\frac{S}{90}=\frac{220S}{130\times 90}=\frac{11S}{585}$

Ratio of time of arrival of B to A

$\frac{t_B}{t_A}=\frac{\frac{11S}{585}}{\frac{S}{55}}=\frac{121}{117}=1.034$

B will take 1.034 times the time of A from Boston to Hartford.

8 0

2 years ago

You are playing a game called "Will It Float?" In this game, you are given a large, square can of tuna. If you know the density

Delicious77 [7]

The only information you would need to decide if the can will float is the density of the can, which requires knowing the mass and volume. If the density of the can is less than one, the can will float. if it is greater than one, it will not float, as water's density is one.

6 0

2 years ago

Read 2 more answers

Springfield's "classic rock" radio station broadcasts at a frequency of 102.1 mhz. what is the length of the radio wave in meter

Mila [183]

The frequency of the radio wave is:
$f=102.1 MHz = 102.1 \cdot 10^6 Hz$

The wavelength of an electromagnetic wave is related to its frequency by the relationship
$\lambda= \frac{c}{f}$
where c is the speed of light and f the frequency. Plugging numbers into the equation, we find
$\lambda= \frac{3 \cdot 10^8 m/s}{102.1 \cdot 10^6 Hz}= 2.94 m$
and this is the wavelength of the radio waves in the problem.

7 0

2 years ago

Two speakers, A and B, produce identical sound waves. A listener is 3.2 m away from speaker A. The listener finds the lowest fre

earnstyle [38]

Answer:

0.83 m or 5.57 m

Explanation:

Destructive interference will occur when the distances from the speakers differ by 1/2 wavelength.

The length of 1 cycle of 72.4 Hz is ...

λ = v/f = (343 m/s)/(72.4 Hz) ≈ 4.738 m

So, the distance of the listener from speaker B is ...

3.2 m ± (4.738 m)/2 = {0.83 m, 5.57 m} . . . either of these distances

_____

The location could be at additional multiples of 4.738 m, but we think not. The sound intensity drops off with the square of the distance from the speaker, so identical sound waves from the speakers will sound quite different at different distances from the speakers. For best interference, the distances need to be as close to the same as possible. That will be at 3.2 m and 5.57 m.

_____

<em>Comment on the speed of sound</em>

We don't know what speed you are to use for the speed of sound. We have used 343 m/s. Some sources use 340 m/s, which will give a result different by 2 or 3 cm.

8 0

2 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

Blizzard [7]

<h2>Answer: 117.626m/s</h2>

Explanation:

The escape velocity $V_{esc}$ is given by the following equation:

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

Where:

$G$ is the Gravitational Constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$

$M$ is the mass of the asteroid

$R$ is the radius of the asteroid

On the other hand, we know the density of the asteroid is $\rho=3.84(10)^{8}g/m^{3}$ and its volume is $V=2.17(10)^{12}m^{3}$ .

The density of a body is given by:

$\rho=\frac{M}{V}$ (2)

Finding $M$ :

$M=\rhoV=(3.84(10)^{8} g/m^{3})(2.17(10)^{12}m^{3})$ (3)

$M=8.33(10)^{20}g=8.33(10)^{17}kg$ (4) This is the mass of the spherical asteroid

In addition, we know the volume of a sphere is given by the following formula:

$V=\frac{4}{3}\piR^{3}$ (5)

Finding $R$ :

$R=\sqrt[3]{\frac{3V}{4\pi}}$ (6)

$R=\sqrt[3]{\frac{3(2.17(10)^{12}m^{3})}{4\pi}}$ (7)

$R=8031.38m$ (8) This is the radius of the asteroid

Now we have all the necessary elements to calculate the escape velocity from (1):

$V_{esc}=\sqrt{\frac{2(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(8.33(10)^{17}kg)}{8031.38m}}$ (9)

Finally:

$V_{esc}=117.626m/s$ This is the minimum initial speed the rocks need to be thrown in order for them never return back to the asteroid.

6 0

2 years ago