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4 years ago

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8) In a bumper car arena, two cars of equal mass are heading straight toward each other. The orange car is traveling at a veloci

ty of 5 m s . The green car is traveling at a velocity of 2 m s . After the collision, both cars move off in the direction in which the orange car was travelling with a velocity of 1.5 m s . Which car has the largest change in velocity?
A) green, because its velocity was slower
B) green, because it changed direction
C) orange, because it changed direction
D) neither, because they both change equally

1 answer:

Ghella [55]4 years ago

3 0

The answer is D) Niether . I just took the test

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Consider a steel guitar string of initial length L=1.00 meter and cross-sectional area A=0.500 square millimeters. The Young's m

Reil [10]

Answer:

$\Delta l=0.015m$

Explanation:

We have given initial length of the steel guitar l = 1 m

Cross sectional area $A=0.5mm^2=0.5\times 10^{-6}m^2$

Young's modulus $\gamma=2\times 10^{11}Pa$

Force F = 1500 N

So stress $=\frac{force}{area}=\frac{1500}{0.5\times 10^{-6}}=3000\times 10^{-6}=3\times 10^{9}Pa$

We know that young's modulus $=\frac{stress}{strain}$

So $2\times 10^{11}=\frac{3\times 10^{9}}{strain}$

$strain=1.5\times 10^{-2}=0.015m$

Now strain $=\frac{\Delta l}{l}$

$0.015=\frac{\Delta l}{1}$

$\Delta l=0.015m$

6 0

3 years ago

Read 2 more answers

There are two parallel conductive plates separated by a distance d and zero potential. Calculate the potential and electric fiel

taurus [48]

Answer:

The total electric potential at mid way due to 'q' is $\frac{q}{4\pi\epsilon_{o}d}$

The net Electric field at midway due to 'q' is 0.

Solution:

According to the question, the separation between two parallel plates, plate A and plate B (say) = d

The electric potential at a distance d due to 'Q' is:

$V = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d}$

Now, for the Electric potential for the two plates A and B at midway between the plates due to 'q':

For plate A,

$V_{A} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}$

Similar is the case with plate B:

$V_{B} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}$

Since the electric potential is a scalar quantity, the net or total potential is given as the sum of the potential for the two plates:

$V_{total} = V_{A} + V_{B} = \frac{1}{4\pi\epsilon_{o}}.q(\frac{1}{\frac{d}{2}} + \frac{1}{\frac{d}{2}}$

$V_{total} = \frac{q}{4\pi\epsilon_{o}d}$

Now,

The Electric field due to charge Q at a distance is given by:

$\vec{E} = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d^{2}}$

Now, if the charge q is mid way between the field, then distance is $\frac{d}{2}$ .

Electric Field at plate A, $\vec{E_{A}}$ at midway due to charge q:

$\vec{E_{A}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}$

Similarly, for plate B:

$\vec{E_{B}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}$

Both the fields for plate A and B are due to charge 'q' and as such will be equal in magnitude with direction of fields opposite to each other and hence cancels out making net Electric field zero.

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3 years ago

A piece of wood 400N in weight and 50cm X 30cm X 20cm in size lies on 50cm X 20cm face. Calculate the pressure exerted by it.

Serga [27]

Answer:

P = 4000 [Pa]

Explanation:

Pressure is defined as the relationship between Force and the area where the body rests.

The support area is equal to:

$A=50*20=1000[cm^{2} ]$

But we must convert from square centimeters to square meters.

$1000[cm^{2}]*\frac{1^{2}m^{2} }{100^{2}m^{2} }=0.1[m^{2} ]$

And the pressure is:

$P=\frac{F}{A} \\P=400/0.1\\P=4000[N/m^{2} ]or 4000[Pa]$

6 0

3 years ago

An object has a mass of 30 grams and measures 3cmx2cmx1cm. What is the density of the object?

dimulka [17.4K]

The density of the object is the ratio of its mass and volume. From the given dimensions above, we determine the volume through the equation,

V = L x W x H

Substituting,

V = (3 cm)(2 cm)(1 cm) = 6 cm³

From the idea presented above,
d = m/V

Substituting the known values,

d = (30 g)/ (6 cm³) = 5 g/cm³

ANSWER: 5 g/cm³

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3 years ago

During a very quick stop, a car decelerates at 7.6 m/s2. Assume the forward motion of the car corresponds to a positive directio

Mashcka [7]

Answer:

24.57 revolutions

Explanation:

(a) If they do not slip on the pavement, then the angular acceleration is

$\alpha = a / r = 7.6 / 0.26 = 29.23 rad/s^2$

(b) We can use the following equation of motion to find out the angle traveled by the wheel before coming to rest:

$\omega^2 - \omega_0^2 = 2\alpha\Delta \theta$

where v = 0 m/s is the final angular velocity of the wheel when it stops, $\omega_0$ = 95rad/s is the initial angular velocity of the wheel, $\alpha = -29.23 rad/s^2$ is the deceleration of the wheel, and $\Delta \theta$ is the angle swept in rad, which we care looking for:

$0 - 95^2 = 2*29.23\Delta \theta$

$9025 = 58.46 \Delta \theta$

$\Delta \theta = 9025 / 58.46 = 154.375 rad$

As each revolution equals to 2π, the total revolution it makes before stop is

154.375 / 2π = 24.57 revolutions

8 0

3 years ago