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ziro4ka [17]
1 year ago
12

A toy car is given an initial velocity of 5.0 m/s and experiences a constant acceleration of 2.0 m/s656-03-02-00-00_files/i02900

00.jpg. What is the final velocity after 6.0 s?
Physics
2 answers:
Anvisha [2.4K]1 year ago
7 0

Answer:

Final velocity of the car, v = 17 m/s

Explanation:

It is given that,

Initial velocity of the car, u = 5 m/s

Acceleration of the car, a=2\ m/s^2

Time, t = 6 s

We need to find the final velocity of the car after 6 seconds. It can be calculated using first equation of motion as :

v=u+at, v is the final velocity of the car

v=5+2\times 6

v = 17 m/s

So, the final velocity of the car is 17 m/s. Hence, this is the required solution.

Whitepunk [10]1 year ago
5 0
It would be 17 m/s

If we use

V2 = V1 + a*t
Sub in 5 for v1
2m/s*2 for a
And
6 for t
That should give you the answer.
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10. A girl pulls a wagon along a level path for a distance of 44 m. The handle of
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The work done on the wagon is 3549 J

Explanation:

The work done by a force when moving an object is given by

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where :

F is the magnitude of the force

d is the displacement

\theta is the angle between the direction of the force and of the displacement

In this problem we have the following data:

F = 87 N is the magnitude of the force

d = 44 m is the displacement of the wagon

\theta=22^{\circ} is the angle between the direction of the force and the displacement

Substituting, we find the work done

W=(87)(44)(cos 22^{\circ})=3549 J

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4 0
2 years ago
The weights of a large number of miniature poodles are approximately normally distributed with a mean of 99 kilograms and a stan
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Answer:

See explanation below

Explanation:

As I say in the comments, the question is incomplete, however, I will try to answer this by using data that I found on another site.

This is the part of the question that is not here:

If measurements are recorded to the nearest

tenth of a kilogram, find the fraction of these poodles

with weights

(a) over 9.5 kilograms;

(b) of at most 8.6 kilograms;

So, assuming a mean of 8 kg, and 0.9 of standard deviation, let X represents the weight of the poodles

The expression to calculate the fraction of poodle needed is:

Z = X - u / d

u: weight of the large number of poodle

d: standard deviation

Replacing data of a) wer have:

Z = 9.5 - 8 / 0.9

Z = 1.67

With this value, we need to take the value of Z, and see the area under the curve of standard deviation (see table attached)

Therefore:

P (X > 9.5) = P(Z > 1.67) = 0.5 - P (Z < 1.67) = 0.5 - 0.4525 = 0.0475

b) In this part, is the same as part a) so:

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The value for area in the curve is 0.2486 so:

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8 0
2 years ago
A majorette in a parade is performing some acrobatic twirlings of her baton. Assume that the baton is a uniform rod of mass 0.12
den301095 [7]

Question

Initially, the baton is spinning about a line through its center at angular velocity 3.00 rad/s.  What is its angular momentum? Express your answer in kilogram meters squared per second.

Answer:

0.0192 kgm^{2}/s

Explanation:

The angular momentum L of the baton moving about an axis perpendicular to it, passing through the center of the baton is,

L = \frac{1}{{12}}m{l^2}\omega

Here, l is the length of the baton.

Substitute 0.120 kg for m, 3 rads/s for \omega[\tex] and 0.8 m for l [tex]\begin{array}{c}\\L = \frac{1}{{12}}m{l^2}\omega \\\\ = \frac{1}{{12}}\left( {0.120{\rm{ kg}}} \right){\left( {{\rm{80}}{\rm{.0 cm}}} \right)^2}{\left( {\frac{{1 \times {{10}^{ - 2}}{\rm{m}}}}{{1{\rm{ cm}}}}} \right)^2}\left( {{\rm{3}}{\rm{.00 rad/s}}} \right)\\\\ = 0.0192{\rm{ kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/s}}\\\end{array}

5 0
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Read 2 more answers
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Fortunately, we don't need to worry at all about the steps in order to derive a first approximation to the answer ... one that's certainly good enough for high school Physics.

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During a family trip to Laura's grandmother's house, the family cast traveled a distance of of 8 miles in 24 minutes. During the
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The correct answer is B

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