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2 years ago

A wagon has a mass of 100 grams and an

Physics

1 answer:

Elza [17]2 years ago

7 0

Answer: 0.2N

Explanation:

mass of wagon = 100 grams

Convert mass in grams to kilograms

If 1000 grams = 1kg

100 grams = (100/1000) = 0.1kg

Acceleration of wagon = 2m/s/s

Force on the wagon = ?

Recall that Force is the product of mass of an object by the acceleration by which it moves.

i.e Force = mass x acceleration

Force = 0.1kg x 2m/s/s

Force = 0.2N

Thus, the force on the wagon is 0.2Newton

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What influences the strength of an electric field?

Slav-nsk [51]

Answer:

Explanation:

The charge alters that space, causing any other charged object that enters the space to be affected by this field. The strength of the electric field is dependent upon how charged the object creating the field is and upon the distance of separation from the charged object.

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3 years ago

How to solve 30(a)<br><br> Give solution asap.

expeople1 [14]

Answer:

They will not meet

h-hX=1.2*g*t²

hX=v0*t-(1/2*g*t²)

Explanation:

fall h=1/2*g*t²

elevation time if v0=20 m/s te=v0/g=20 m/s /9.81 m/s²=2.0387s

hmax=v0²/(2*g)=(400 m²/s²)/19.62 m/s²2=20.387 m

free fall

t=2.0387s yields hX=1/2*g*t²=20.387 m

h-hX=200m - 20.387 m=179,613 m.

so, the second body has not enough initianoal speed to reach a meeting point

5 0

3 years ago

When potential energy is converted into Kinetic energy, some of that kinetic energy is always in the form of

Paraphin [41]

Answer: thermal energy

Explanation: Thermal energy is the conversion of kinetic energy

7 0

2 years ago

If the spring constant k of a pogo stick is 3500 N m and the weight of the person on the pogo stick is 700 N, how much is the sp

algol [13]

As we know that spring force is given by

$F = kx$

here we know that

F = 700 N

also we know that

k = 3500 N/m

now from above equation we will have

$700 = 3500 (x)$

now we will have

$x = \frac{700}{3500}$

$x = \frac{1}{5} m = 20 cm$

so here we can say that end of the spring will be compressed by 20 cm

6 0

2 years ago

A child is sliding a toy block (with mass = m) down a ramp. The coefficient of static friction between the block and the ramp is

tiny-mole [99]

Answer:

$F=mg(sin(\theta )-0.25 cos(\theta ))$

Explanation:

The free body diagram of the block on the slide is shown in the below figure

Since the block is in equilibrium we apply equations of statics to compute the necessary unknown forces

N is the reaction force between the block and the slide

For equilibrium along x-axis we have

$\sum F_{x}=0\\\\mgsin(\theta )-\mu N-F=0\\\therefore F=mgsin(\theta)-\mu N......(\alpha )\\Similarly\\\sum F_{y}=0\\\\N-mgcos(\theta )=0\\\therefore N=mgcos(\theta ).......(\beta )\\\\$

Using value of N from equation β in α we get value of force as

$F=mg(sin(\theta )-\mu cos(\theta ))$

Applying values we get

$F=mg(sin(\theta )-0.25 cos(\theta ))$

8 0

2 years ago

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