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s344n2d4d5 [400]

2 years ago

10

A shovel is the third class lever

1 answer:

Nookie1986 [14]2 years ago

8 0

Answer:

Yes

Explanation:

In a third-class lever, the effort force lies between the resistance force and the fulcrum. Some kinds of garden tools are examples of third-class levers. When you use a shovel, for example, you hold one end steady to act as the fulcrum, and you use your other hand to pull up on a load of dirt.

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Show all work please I am stuck

MissTica

Answer:

Explanation:

1 meter = 39.37 inches

meters x inches/meters = inches

6.23x10^-4 m x 39.37 in./m=0.245...=2.45x10^-2

5 0

2 years ago

Two particles, each of mass m, are initially at rest very far apart.Obtain an expression for their relative speed of approach at

PSYCHO15rus [73]

Answer:

$|\Delta v |=\sqrt{\frac{4Gm}{d} }$

Explanation:

Consider two particles are initially at rest.

Therefore,

the kinetic energy of the particles is zero.

That initial K.E. = 0

The relative velocity with which both the particles are approaching each other is Δv and their reduced masses are

$\mu= \frac{m_1m_2}{m_1+m_2}$

now, since both the masses have mass m

therefore,

$\mu= \frac{m^2}{2m}$

= m/2

The final K.E. of the particles is

$KE_{final}=\frac{1}{2}\times \mu\times \Delta v^2$

Distance between two particles is d and the gravitational potential energy between them is given by

$PE_{Gravitational}= \frac{Gmm}{d}$

By law of conservation of energy we have

$KE_{initial}+KE_{final}= PE_{gravitaional}$

Now plugging the values we get

$0+\frac{1}{2}\frac{m}{2}\Delta v^2= -\frac{Gmm}{d}$

$|\Delta v |=\sqrt{\frac{4Gm}{d} }$

$=\sqrt{\frac{Gm}{d} }$

This the required relation between G,m and d

5 0

2 years ago

A thin ring of radius 73 cm carries a positive charge of 610 nC uniformly distributed over it. A point charge q is placed at the

kow [346]

Answer:

q = - 93.334 nC

Explanation:

GIVEN DATA:

Radius of ring 73 cm

charge on ring 610 nC

ELECTRIC FIELD p FROM CENTRE IS AT 70 CM

E = 2000 N/C

Electric field due tor ring is guiven as

$E = \frac{KQx}{[x^2+ R^2]^{3/2}}$

$E = \frac{9\time 10^9 \times 610\times 10^[-9} 0.70}{(0.70^2 + 0.73^2)^{3/2}}$

E1 = 3714.672 N/C

electric field due to point charge q

$E =\frac[kq}{x^2}$

$E = \frac{9\times 10^9 \times q}{0.70^2}$

$E2 = 1.837\times 10^{10}\times q$

now the eelctric charge at point P is

E = E1 + E2 $2000 = 3714.672 + 1.837\times 10[10} \times q$

solving for q

q = - 93.334 nC

7 0

3 years ago

Please select the word from the list that best fits the definition

vagabundo [1.1K]

Answer:input work

Explanation:

7 0

2 years ago

Read 2 more answers

Whenever Dylan walks into his room, his roommate loudly blows an air horn, which startles Dylan. Dylan doesn't think this is fun

Sav [38]

Answer:

stimuli

Explanation:

4 0

2 years ago