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Georgia [21]
3 years ago
10

URGENT!! Question in attachment. thanks ​

Physics
2 answers:
Roman55 [17]3 years ago
6 0

Answer:9

Explanation:beacause

jarptica [38.1K]3 years ago
6 0

Answer:

a) describe, step by step, what happens

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3 years ago
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How to convert 8 feet to metres​
ohaa [14]

Answer:

2.4384m

Explanation:

1 ft in m is 0.3048

8ft in m is x

therefore 8ft in m is

8 * 0.3048 = 2.4384

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Draw velocity-ime graph for uniform moion of an object , when initially body is at rest.
Klio2033 [76]
When the body is at rest, its speed is zero, and the graph lies on the x-axis.

When the body is in uniform motion, the speed is constant, and the graph is a horizontal line, parallel to the x-axis and some distance above it.

It's impossible to tell, based on the given information, how these two parts of the
graph are connected.  There must be some sloping (accelerated) portion of the graph
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8 0
3 years ago
A force of 4 kg weight acts on a body of mass 9.8 kg calculate the acceleration
White raven [17]

Answer:

here given is a weight

then force becomes mg

that is F=Mg

=4*9.8

then by using the formula

F=Ma

a=F/M

=4*9.8/9.8

=4

Explanation:

3 0
3 years ago
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A car is moving at 19 m/s along a curve on a horizontal plane with radius of curvature 49m.
JulsSmile [24]

Answer:

\mu =0.75

Explanation:

<u>Frictional Force </u>

When the car is moving along the curve, it receives a force that tries to take it from the road. It's called centripetal force and the formula to compute it is:

F_c=m.a_c

The centripetal acceleration a_c is computed as

\displaystyle a_c=\frac{v^2}{r}

Where v is the tangent speed of the car and r is the radius of curvature. Replacing the formula into the first one

F_c=m.\frac{v^2}{r}

For the car to keep on the track, the friction must have the exact same value of the centripetal force and balance the forces. The friction force is computed as

F_r=\mu N

The normal force N is equal to the weight of the car, thus

F_r=\mu .m.g

Equating both forces

\displaystyle \mu .m.g=m.\frac{v^2}{r}

Simplifying

\displaystyle \mu =\frac{v^2}{rg}

Substituting the values

\displaystyle \mu =\frac{19^2}{(49)(9.8)}

\boxed{\mu =0.75}

7 0
3 years ago
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