What prevents geologists from exploring earth's interior?

The statement which describes how a machine can help make work easier is that It can put out more force than the input force by decreasing the distance over which force is applied, therefore the correct option is option A.

The total amount of energy transferred when a force is applied to move an object through some distance

The work done is the multiplication of applied force with displacement.

Work Done = Force * Displacement

As work done depends both on the force as well as the displacement force and be reduced by reducing the displacement if the same amount of work is performed by the machine.

The correct answer is option A since the statement that describes how a machine might assist in making work easier says that it can put out greater force than the input force by reducing the distance over which force is exerted.

Learn more about work done from here

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5 0

2 years ago

A student, along with her backpack on the floor next to her, are in an elevator that is accelerating upward with acceleration a.

Anna007 [38]

Answer:

$\mu_k = \frac{2(vt - L)}{(g + a) t^2}$

Explanation:

As we know that backpack is kicked on the rough floor with speed "v"

So here as per force equation in vertical direction we know that

$N - mg = ma$

so normal force on the block is given as

$N = mg + ma$

now the magnitude of kinetic friction on the block is given as

$F_f = \mu N$

$F_f = \mu (mg + ma)$

now when bag is sliding on the floor then net deceleration of the block due to friction is given as

$a = - \frac{F_f}{m}$

$a = -\mu_k(g + a)$

now we know that bag hits the opposite wall at L distance away in time t

so we have

$d = v t + \frac{1}{2}at^2$

$L = vt - \frac{1}{2}(\mu_k)(g + a) t^2$

$\mu_k = \frac{2(vt - L)}{(g + a) t^2}$

8 0

3 years ago

Introduction: The specific heat capacity of a substance is the amount of energy needed to change the temperature of that substan

ki77a [65]

Answer:

A) 8,368 J

B) ) 0.893 J/gºC

Explanation:

A)

The heat gained by the water can be obtained solving the following equation:

$q_{g} = c_{w} * m * \Delta T (1)$

where cw = specific heat of water = 4.184 J/gºC
m= mass of water = 1,000 g
ΔT = 2ºC
Replacing these values in (1) we get:

$q_{g} = c_{w} * m * \Delta T = 4.184 J/gºC*1,000 g* 2ºC = 8,368 J (2)$

B)

Assuming that the heat energy gained by the water is equal to the one lost by the aluminum, we can use the same equation, taking into account that the energy is lost by the aluminum, so the sign is negative: -8,368 J.
Replacing by the mass of aluminum (125 g), and the change in temperature (-74.95ºC), in (1), we can solve for the specific heat of aluminum, as follows:

$q_{l} = c_{Al} * m_{Al} * \Delta T (3)$

⇒ $-8,368 J = c_{Al}* 125 g * (-74.95ºC) (4)$

$c_{Al} = \frac{-8,368J}{125g*(-74.95ºC} = 0.893 J/gºC (5)$

which is pretty close to the Aluminum's accepted specific heat value of 0.900 J/gºC.

8 0

3 years ago