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natka813 [3]
2 years ago
12

Denise is conducting a physics experiment to measure the acceleration of a falling object when it slows down and comes to a stop

. She drops a wooden block with a mass od 0.5 kilograms on a sensor on the floor. The sensor measures the force of the impact as 4.9 newtons. Whats the acceleration of the wooden block wgen it hits the sensor
Physics
2 answers:
iren [92.7K]2 years ago
4 0
We need a and we have m and F . Now a = f÷m so therefore a = 4,9 ÷ 0,5 which is 0,98 metres per cubic second
Roman55 [17]2 years ago
3 0

Answer:

a = 9.8 m/s^2

Explanation:

As we know that here given parameters are

mass = 0.5 kg

also we know that force of impact on the floor is given as

F = 4.9 N

now we know by the Newton's 2nd law says that the net force on an object is product of mass and its acceleration

it is given by

F = ma

now we have

a = \frac{F}{m}

a = \frac{4.9}{0.5}

a = 9.8 m/s^2

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Two parallel wires carry a current I in the same direction. Midway between these wires is a third wire, also parallel to the oth
Luda [366]

Answer:

Force is repulsive hence direction of force is away from wire

Explanation:

The first thing will be to draw a figure showing the condition,

Lets takeI attractive force as +ve and repulsive force as - ve and thereafter calculating net force on outer left wire due to other wires, net force comes out to be - ve which tells us that force is repulsive, hence direction of force is away from wire as shown in figure in the attachment.

4 0
2 years ago
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Two particles carrying charges q1 and q2 are separated by a distance r and exert an electric force F⃗ E on each other. If q1 is
zepelin [54]

Answer:

q2 must also be doubled

r may also be halved

Explanation:

According to Coulumbs law

F= K q1 q2/r^2

If q1 is doubled, we must necessarily double q2 and r may also be halved in order to maintain F at the same value. Once the value of F is thus kept constant and E is also constant, the product FE must remain constant.

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2 years ago
A man climbs a ladder. Which two quantities can be used to calculate the energy stored of the man at the top of the ladder.
Dvinal [7]

Answer:The answer must be The weight of the man and the vertical distance moved.

Explanation: you calculate it by the force you applied times the distance you moved

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2 years ago
What is the displacement of a runner who is running at 0.5km/h for 3 hours due north
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I am so sorry but I am not sure if the answers I have is accurate
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2 years ago
A charged box (m=445 g, ????=+2.50 μC) is placed on a frictionless incline plane. Another charged box (????=+75.0 μC) is fixed i
victus00 [196]

The concept required to perform this exercise is given by the coulomb law.

The force expressed according to this law is given by

F= \frac{kqQ}{r^2}

Where,

k = 8.99 * 10^9 N m^2 / C^2.

q = charges of the objects

r= distance/radius

Our values are previously given, so

q= 2.5*10^{-6}C\\Q= 75*10^{-6}C\\r=0.59

Replacing,

F=\frac{kqQ}{r^2}

F= \frac{(8.99 x 10^9)(2.5*10^{-6})(75*10^{-6})}{0.59^2}

F= 4.8423N

The force acting on the block are given by,

F-mgsin\theta = ma

a = \frac{F-mgsin\theta}{m}

a = \frac{4.8423-(0.445)(9.8)sin(35)}{0.445}a = 10.31m/s^2

Therefore the box is accelerated upward.

3 0
2 years ago
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