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wolverine [178]
2 years ago
15

A force is applied to an object at rest with a mass of 100kg. A force twice as large is applied to another object at rest with a

mass of 100kg. What differences in their motion would you observe? Explain your reasoning.
Physics
1 answer:
sukhopar [10]2 years ago
7 0
According to Newton's second law. £Fext=m.a
then the acceleration of the second object will be twice as that of the first since the mass of both objects is equal(100kg) and the force F is doubled. Therefore the acceleration of object 2 will be more than that of object 1 (double)

hope this helps.

best regards
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4. Una cuerda de acero de piano mide 1.60 m de longitud y 0.20 cm de diámetro. ¿Cuál es la tensión en la cuerda si se estira 0.2
bazaltina [42]

Answer:

1030.83\ \text{N}

Explanation:

\Delta L = Cambio en la longitud de la cuerda = 0.25 cm

T = tensión en cuerda

A = Área de la cadena = \dfrac{\pi}{4}d^2

d = Diámetro de la cuerda = 0.2 cm

L = Longitud original de la cuerda = 1.6 m

El cambio de longitud de una cuerda viene dado por

\Delta L=\dfrac{TL}{AE}\\\Rightarrow T=\dfrac{\Delta LAE}{L}\\\Rightarrow T=\dfrac{0.25\times 10^{-2}\times \dfrac{\pi}{4}(0.2\times 10^{-2})^2\times 210\times 10^9}{1.6}\\\Rightarrow T=1030.84\ \text{N}

La tensión en la cuerda es 1030.84\ \text{N}.

8 0
2 years ago
A basketball player jumped straight up to grab a rebound. If she was in the air for 0.80 second, how high did she jump?
Kazeer [188]
-- She went up for 0.4 sec and down for 0.4 sec.

-- The vertical distance traveled in gravity during ' t ' seconds is

                   D  =  (1/2)  x  (g)  x  (t)²

                       = (1/2) (9.8 m/s²) (0.4 sec)²

                       =    (4.9 m/s²)  x  (0.16 s²)

                       =      0.784 meter        ( B )
4 0
3 years ago
Read 2 more answers
Q11) If you were standing at the top of a building and you dropped a rock.
Dafna1 [17]

Answer:

Part A

The distance travel by the rock is approximately 132.496 m

Part B

The speed when the rock hits the ground is approximately 50.96 m/s

Explanation:

Part A

The question is focused on the kinetics equation of a free falling object

The given parameter is the time it takes the rock to hit the ground, t = 5.2 s

For an object in free fall, we have;

h = 1/2·g·t²

Where;

h = The height from which the object is dropped

g = The acceleration due to gravity ≈ 9.8 m/s²

t = The time taken to travel the distance, h = 5.2 s

∴ h = 1/2 × 9.8 m/s² × (5.2 s)² ≈ 132.496 m

The distance travel by the rock, h ≈ 132.496 m

Part B

The speed, 'v', when the rock hits the ground, is given by the following kinematic equation,

v = g·t

∴ v = 9.8 m/s² × 5.2 s = 50.96 m/s

The speed when the rock hits the ground, v ≈ 50.96 m/s.

8 0
2 years ago
A 1.50 µF capacitor and a 3.50 µF capacitor are connected in series across a 2.50 V battery. How much charge (in µC) is stored o
Nataly_w [17]

Explanation:

The given data is as follows.

      C_{1} = 1.50 \times 10^{-6} F

      C_{1} = 3.50 \times 10^{-6} F    

      Voltage = 2.50 V

Hence, calculate the equivalence capacitor as follows.

    \frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}}

    \frac{1}{C} = \frac{1}{1.50 \times 10^{-6} F} + \frac{1}{3.50 \times 10^{-6} F}

                 = 0.945 \times 10^{-6} F

          C = 1.06 \times 10^{-6} F

Now, we will calculate the charge across each capacitance as follows.

              Q = CV

                  = 1.06 \times 10^{-6} F \times 2.50 V

                  = 2.65 \times 10^{-6} C

                  = 2.65 \mu C

Thus, we can conclude that 2.65 \mu C is the charge stored on each given capacitor.

5 0
3 years ago
If an object is accelerating to the right , then net force on the object must be directed towards the right ! True Or False ?
vichka [17]
Not so fast.

I think you're using 'accelerating' to mean 'speeding up', but you really need
to be more careful with it. "Acceleration" means ANY change in speed OR
direction.

If an object's speed to the left is decreasing, or its speed to the right is
increasing, then the net force on the object must be directed towards
the right.

If an object is moving with constant speed in a circular path, then it's
constantly accelerating, because its direction is constantly changing.
The force on it is always directed towards the center of the circle, so
there's one point on the path where the force is directed straight to the right.
8 0
2 years ago
Read 2 more answers
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