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horrorfan [7]
3 years ago
7

Elizabeth, with a mass of 56.1kg stands on a scale in an elevator. Total mass of elevator plus Elizabeth=850kg. As the elevator

starts moving, the scale reads 450N. Find the acceleration and tension in the cable.
Physics
1 answer:
Reil [10]3 years ago
7 0

Answer:

acceleration = 1.79 m/s^2

Tension = 6817 N

Explanation:

First let's find Elizabeth's weight:

P = m*g = 56.1 * 9.81 =  550.34\ N

Her weight is greater than the normal force (N = 450 N), so the elevator is going downwards.

The acceleration of Elizabeth is given by:

P - N = m*a

Where P is the weight of Elisabeth, N is her normal force, m is her mass and a is the acceleration. Then, we have that:

56.1*9.81 - 450 = 56.1*a

a = 100.34 / 56.1 = 1.79\ m/s^2

The tension in the cable is given by:

P - T = m*a

In this case, we use the total mass, so we have:

850*9.81 - T = 850*1.79

T = 850 * 8.02 = 6817\ N

You might be interested in
A 96.0 kg hoop rolls along a horizontal floor so that its center of mass has a speed of 0.240 m/s. How much work must be done on
Ede4ka [16]

Answer:

The work done by the hoop is equal to 5.529 Joules.

Explanation:

Given that,

Mass of the hoop, m = 96 kg

The speed of the center of mass, v = 0.24 m/s

To find,

The work done by the hoop.

Solution,

The initial energy of the hoop is given by the sum of linear kinetic energy and the rotational kinetic energy. So,

K_i=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2

I is the moment of inertia, I=mr^2

Since, \omega=\dfrac{v}{r}

K_i=mv^2

K_i=96\times (0.24)^2=5.529\ J

Finally it stops, so the final energy of the hoop will be, K_f=0

The work done by the hoop is equal to the change in kinetic energy as :

W=K_f-K_i

W = -5.529 Joules

So, the work done by the hoop is equal to 5.529 Joules. Therefore, this is the required solution.

4 0
3 years ago
Dibuja la gráfica de calentamiento de un kilogramo de plomo que se encuentra inicialmente a 70ºC y pasa a una temperatura final
MariettaO [177]

Answer:

Q= m c_e ΔT and   Q = m L

Explanation:

For this graph of temperature vs energy (heating) we must use two relations

* for when there is no change of state

          Q= m c_e ΔT

* for using there is change of state

          Q = m L

the second expression is a consequence of the fact that all the energy supplied is used to change the state of the solid-liquid and liquid-gas system

the energy supplied is the sum of the energy in each interval

divide the system into intervals determined by the state change points

1) from T₀ = 70ºC to T_f = 327.4ºC, sample in solid-liquid state

           c_e = 128 J / kg ºC

           Q₁ = m c_e (T_f -To)

           Q₁=1  128 (327.4 -70)

           Q₁ = 3.29 10⁴ J

           Q = Q₁ = 3.29 10⁴ J

2) when is it changing from solid to liquid

            L = 2.45 10⁴ J / kg

            Q2 = 1 2.45 10⁴

            Q2 = 2.45 10⁴ J

            Q = Q₁ + Q₂

             Q = 5.74 10⁴ J

3) from to = 327.4ºC until T_f = 1725ºC, sample in liquid state

in the tables the specific heat of the solid and liquid state is the same

             Q3 = m c_e (T_f -To)

             Q3 = 1 128 (1725 -327.4)

             Q3 = 1.79 10⁵ J

              Q = Q₁ + Q₂ + Q₃

              Q = (3.29 +2.45 + 17.9) 10⁴ J

              Q = 23.64 10⁴ J

4) for when it is changing from the liquid state to the gaseous state

             L_v = 8.70 10⁵ J / kg

             Q₄ = m L_v

             Q₄ = 1 8.70 10⁵

             Q₄ = 8.70 10⁵ J

             Q = Q₁ + Q₂ + Q₃ + Q₄

              Q = (3.29 +5.74 + 17.9+ 87.0) 10⁴ J

               Q = 110.64 10⁴ J

5) from To = 1725ºC to T_f = 2000ºC, sample in gaseous state

             Q₅ = m c_e ΔT

             Q₅ = 1 128 (2000 -1725)

             Q₅ = 3.52 10⁴ J

             Q = Q₁ + Q₂ + Q₃ + Q₄ + Q₅

              Q = 114.16 104 J

the following table shows the points to be plotted

         Energy (10⁴ J)  Temperature (ºC)

                  0                     70

                 3.29             327.4

                 5.74             327.4

               23.64           1725

               110.64          1725

                114.16         2000

In the attachment we can see a graph of Temperature versus energy supplied

8 0
3 years ago
A force of 1 N is the only horizontal force exerted on a block, and the horizontal acceleration of the block is
SVEN [57.7K]

The mass of the first block will be three times the mass of the second block.

According to Newton's second law of motion, the force acting on a body is directly proportional to the acceleration as shown;

F\ \alpha \ a

F = ma

F is the acting force

m is the mass

a is the acceleration of the body

Given the following parameters

Constant force F =  1N

For the first block with the acceleration of "a"

1 = m₁a

a = m₁/1

m₁ = a .................1

For the second block, acceleration is thrice that of the first. This means;

F = m(3a)

1 = 3ma

m_2=\frac{1}{3a} ..........................2

Divide both equations

\frac{m_1}{m_2} =\frac{a}{(\frac{1}{3a} )}\\\frac{m_1}{m_2} = 3\\m_1 = 3m_2

From the calculation, we can conclude that the mass of the first block will be three times the mass of the second block.

Learn more here: brainly.com/question/19030143

4 0
3 years ago
What would a sound wave do if it traveled from a gas to a
NeTakaya

Answer:

speed up

Explanation:

it's sound, not electromagnetic

5 0
3 years ago
A star rotates with a period of 37 days about an axis through its center. The period is the time interval required for a point o
vlabodo [156]

Answer:

T = 1.1285 10⁻² day

Explanation:

For this exercise the forces in the premiere are internal, so the angular momentum is conserved

         L₀ = I₀ w₀

         L  = I w

         L₀ = L

         I₀ w₀ = I w

Angular velocity and period are related

         w₀ = 2π / T₀

         w = 2π  / T

         

The moment of inertia of a sphere is

       I₀ = 2/5 M R²

       I = 2/5 m r²

If we assume that the mass of the star does not change in the transformation

We substitute

         2/5 M R² 2π/T₀ = 2/5 M r² 2π/ T

          R² /T₀ = r² / T

          T = (r / R)² T₀

          T = (6.1 / 2.0 104) 37

          T = 1.1285 10⁻² day

5 0
3 years ago
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