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DaniilM [7]
2 years ago
7

PLEASE HELP SOLVE THIS PHOTO BELOW

Physics
2 answers:
polet [3.4K]2 years ago
8 0

Answer:

C

Explanation:

The pattern is adding .5 to the cm every .1 in weight you just continue the table

notsponge [240]2 years ago
7 0

Answer:

C

Explanation:

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In a series circuit with three bulbs,
elena-14-01-66 [18.8K]

In a series circuit, there is only one path for current to take. 
If more bulbs are added, then the same current loses more energy,
making heat and light on its way through more bulbs, so the ones that
were there before become dimmer.

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3 years ago
F all of the energy in a falling object's gravitational potential energy store is transferred to its kinetic energy store by the
stepladder [879]

Answer:

The options are not shown, so let's derive the relationship.

For an object that is at a height H above the ground, and is not moving, the potential energy will be:

U = m*g*H

where m is the mass of the object, and g is the gravitational acceleration.

Now, the kinetic energy of an object can be written as:

K = (1/2)*m*v^2

where v is the velocity.

Now, when we drop the object, the potential energy begins to transform into kinetic energy, and by the conservation of the energy, by the moment that H is equal to zero (So the potential energy is zero) all the initial potential energy must now be converted into kinetic energy.

Uinitial = Kfinal.

m*g*H = (1/2)*m*v^2

v^2 = 2*g*H

v = √(2*g*H)

So we expressed the final velocity (the velocity at which the object impacts the ground) in terms of the height, H.

5 0
2 years ago
Gggcdfubrdegubtcwftvf y day rx u e HHS’s forget h
aev [14]

Answer:

um . . . yes ?

8 0
2 years ago
Read 2 more answers
Find the acceleration that can result from a net force of 13 N exerted on a 3.6-kg cart. (Note: The unit N/kg is equivalent to m
STatiana [176]

Answer:

a = 3.61[m/s^2]

Explanation:

To find this acceleration we must remember newton's second law which tells us that the total sum of forces is equal to the product of mass by acceleration.

In this case we have:

F = m*a\\\\m=mass = 3.6[kg]\\F = force = 13[N]\\13 = 3.6*a\\a = 3.61[m/s^2]

7 0
3 years ago
The plug has a diameter of 30 mm and fits within a rigid sleeve having an inner diameter of 32 mm. Both the plug and the sleeve
Katena32 [7]

Answer:

P=740 KPa

Δ=7.4 mm

Explanation:

Given that

Diameter of plunger,d=30 mm

Diameter of sleeve ,D=32 mm

Length .L=50 mm

E= 5 MPa

n=0.45

As we know that

Lateral strain

\varepsilon _t=\dfrac{D-d}{d}

\varepsilon _t=\dfrac{32-30}{30}

\varepsilon _t=0.0667

We know that

n=-\dfrac{\epsilon _t}{\varepsilon _{long}}

\varepsilon _{long}=-\dfrac{\epsilon _t}{n}

\varepsilon _{long}=-\dfrac{0.0667}{0.45}

\varepsilon _{long}=-0.148

So the axial pressure

P=E\times \varepsilon _{long}

P=5\times 0.148

P=740 KPa

The movement in the sleeve

\Delta =\varepsilon _{long}\times L

\Delta =0.148\times 50

Δ=7.4 mm

6 0
3 years ago
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