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3 years ago

Find the height of a baseball with a mass of 0.15 kg that has a GPE of 73.5 J. Help please?

Physics

2 answers:

LUCKY_DIMON [66]3 years ago

8 0

Hight is 50 b/c it is on the ledt side

Slav-nsk [51]3 years ago

4 0

Gravitational Potential Energy of an object, relative to the ground =

   (mass) x (acceleration of gravity) x (height above the ground)

   73.5 J = (0.15 kg) x (9.8 m/s²) x (height above the ground).

1 J = 1 kg · m² / s²

   73.5 kg·m²/s² = (0.15 kg) · (9.8 m/s²) · (height)

Divide each side by (0.15 kg):

   (73.5 kg·m² / (0.15 kg·s²) = (9.8 m/s²) · (height)

Divide each side by (9.8 m/s²):

   (73.5 kg·m²·s²) / (0.15 kg·s² · 9.8 m) = (height)

After doing all the arithmetic on the left side:

   (73.5) / (0.15 · 9.8) (m)   =   height

   50 (m)    =   height

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Select the correct answer.

Dahasolnce [82]

Answer:

a is the correct choice

Explanation:

5 0

2 years ago

Read 2 more answers

A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1

bija089 [108]

Answer:

Part a)

$\rho = 1.35 \times 10^{-5}$

Part b)

$\alpha = 1.12 \times 10^{-3}$

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

$i = 18.7 A$

V = 17.0 volts

now we will have

$V = I R$

$17.0 = 18.7 R$

R = 0.91 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho = 1.35 \times 10^{-5}$

Part b)

Now at higher temperature we have

$V = I R$

$17.0 = 17.3 R$

R = 0.98 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho' = 1.46 \times 10^{-5}$

so we will have

$\rho' = \rho(1 + \alpha \Delta T)$

$1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))$

$\alpha = 1.12 \times 10^{-3}$

Answer:

Part a)

$i = 1.55 A$

Part b)

$v_d = 1.4 \times 10^{-4} m/s$

Explanation:

Part a)

As we know that current density is defined as

$j = \frac{i}{A}$

now we have

$i = jA$

Now we have

$j = 1.90 \times 10^6 A/m^2$

$A = \pi(\frac{1.02 \times 10^{-3}}{2})^2$

so we will have

$i = 1.55 A$

Part b)

now we have

$j = nev_d$

so we have

$n = 8.5 \times 10^{28}$

$e = 1.6 \times 10^{-19} C$

so we have

$1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d$

$v_d = 1.4 \times 10^{-4} m/s$

8 0

3 years ago

child slides down a snow‑covered slope on a sled. At the top of the slope, her mother gives her a push to start her off with a s

Strike441 [17]

Answer:

θ = 13.7º

Explanation:

According to the work-energy theorem, the change in the kinetic energy of the combined mass of the child and the sled, is equal to the total work done on the object by external forces.
The external forces capable to do work on the combination of child +sled, are the friction force (opposing to the displacement), and the component of the weight parallel to the slide.
As this last work is just equal to the change in the gravitational potential energy (with opposite sign) , we can write the following equation:

$\Delta K + \Delta U = W_{nc} (1)$

ΔK, is the change in kinetic energy, as follows:

$\Delta K = \frac{1}{2}* m* (v_{f} ^{2} - v_{0} ^{2}) (2)$

ΔU, is the change in the gravitational potential energy.
If we choose as our zero reference level, the bottom of the slope, the change in gravitational potential energy will be as follows:

$\Delta U = 0 - m*g*h = -m*g*d* sin\theta (3)$

Finally, the work done for non-conservative forces, is the work done by the friction force, along the slope, as follows:

$W_{nc} = F_{f} * d * cos 180\º \\\\ = 0.2*m*g*d* cos 180\º = -0.2*m*g*d (4)$

Replacing (2), (3), and (4) in (1), simplifying common terms, and rearranging, we have:

$\frac{1}{2}* (v_{f} ^{2} - v_{0} ^{2}) = g*d* sin\theta -0.2*g*d$

Replacing by the givens and the knowns, we can solve for sin θ, as follows: $\frac{1}{2}*( (4.30 m/s) ^{2} - (0.75 m/s)^{2}) = 9.8 m/s2*25.5m* sin\theta -0.2*9.8m/s2*25.5m\\ \\ 8.56 (m/s)2 = 250(m/s)2* sin \theta -50 (m/s)2\\ \\ sin \theta = \frac{58.6 (m/s)2}{250 (m/s)2} = 0.236$ ⇒ θ = sin⁻¹ (0.236) = 13.7º

8 0

3 years ago

Assume the ground is uniformly level. If the horizontal component a projectile's velocity is doubled, but the vertical component

Katarina [22]

Answer:

Explanation:

Time of flight = 2 x u sinα / g where u sinα is vertical component of projectile's velocity u .

So Time of flight = 2 x vertical component / g

vertical component = constant

g is also constant so

Time of flight will also be constant .

It will remain unchanged .

3 0

3 years ago

Which statement best describes a device that has a high energy efficiency? Its ratio of energy absorbed to energy released is hi

Stels [109]

Answer:

It wastes little energy

Explanation:

The efficiency of a machine refers to the ratio of work output to the work input as a percentage.
The efficiency of most machines is less than 100% because some work is lost due to friction and heat.
Therefore, a machine with high efficiency has less wastage of energy due to factors such as friction of the moving parts. Conversely, low efficiency means there is more wastage of energy.

3 0

3 years ago

Read 2 more answers