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krek1111 [17]
4 years ago
6

A 7.00 kg bowling ball is held 2.00 m above the ground. Using g = 9.80 m/s2, how much energy does the bowling ball have due to i

ts position?
Physics
2 answers:
Naily [24]4 years ago
5 0

Answer: 137.2 J

Explanation:

The gravitational potential energy of the bowling ball is given by:

U=mgh

where

m = 7.00 kg is the mass of the ball

g = 9.80 m/s^2 is the acceleration due to gravity

h = 2.00 m is the height above the ground

Substituting, we find

U=(7.00 kg)(9.8 m/s^2)(2.00 m)=137.2 J

deff fn [24]4 years ago
4 0
mgh =7*2*9.8=137.2. It should be the answer.
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You need to design a photodetector that can respond to the entire range of visible light. True or False
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Answer: True

Explanation:

A photo detector that can respond to the entire rang of visible light can be design, it is true.

Photo detector is a device in an optical receiver which receives optical signals and convert it to electric signal. It is the key device position in front of the optical receiver.

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3 years ago
Complete the sentence to describe what a wave is and what it does. A wave is a disturbance that carries from one place to anothe
kozerog [31]

Answer:

Energy

A wave is a disturbance that carries energy from one place to another through matter and space.

Explanation:

A wave can be defined as a form of disturbance that carries energy from one place to another through matter and space.

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3 years ago
Read 2 more answers
A rifle bullet with mass ma = 8.00 g strikes and embeds itself in a block with mass mb = 0.992 kg that rests on a frictionless,
Doss [256]

Answer:

the magnitude of the velocity of the block just after impact is 2.598 m/s and the original speed of the bullect is 324.76m/s.

Explanation:

a)  Kinetic energy of block = potential energy in spring  

½ mv² = ½ kx²

Here m stands for combined mass (block + bullet),

which is just 1 kg.  Spring constant k is unknown, but you can find it from given data:  

k = 0.75 N / 0.25 cm

= 3 N/cm, or 300 N/m.  

From the energy equation above, solve for v,

v = v √(k/m)  

= 0.15 √(300/1)

= 2.598 m/s.

b)  Momentum before impact = momentum after impact.

Since m = 1 kg,

v = 2.598 m/s,

p = 2.598 kg m/s.  

This is the same momentum carried by bullet as it strikes the block.  Therefore, if u is bullet speed,  

u = 2.598 kg m/s / 8 × 10⁻³ kg

= 324.76 m/s.

Hence, the magnitude of the velocity of the block just after impact is 2.598 m/s and the original speed of the bullect is 324.76m/s.

6 0
3 years ago
A 1,508 kg car rolling on a horizontal surface has a speed of 20.8 km/hr when it strikes a horizontal coiled spring and is broug
natali 33 [55]

Answer:

Approximately 1.79 \times 10^{5}\; {\rm N \cdot m^{-1}}, assuming friction between the vehicle and the ground is negligible.

Explanation:

Let m denote the mass of the vehicle. Let v denote the initial velocity of the vehicle. Let k denote the spring constant (needs to be found.) Let x denote the maximum displacement of the spring.

Convert velocity of the vehicle to standard units (meters per second):

\begin{aligned}v &= 20.8\; {\rm km \cdot h^{-1}} \times \frac{1000\; {\rm m}}{1\; {\rm km}} \times \frac{1\; {\rm h}}{3600\; {\rm s}} \\ &\approx 1.908\; {\rm m \cdot s^{-1}}\end{aligned}.

Initial kinetic energy ({\rm KE}) of the vehicle:

\begin{aligned}\frac{1}{2}\, m \, v^{2}\end{aligned}.

When the vehicle is brought to a rest, the elastic potential energy (\text{EPE}) stored in the spring would be:

\displaystyle \frac{1}{2}\, k\, x^{2}.

By the conservation of energy, if the friction between the vehicle and the ground is negligible, the initial \text{KE} of the vehicle should be equal to the {\rm EPE} of the vehicle. In other words:

\begin{aligned}\frac{1}{2}\, m \, v^{2} &= \frac{1}{2}\, k\, x^{2}\end{aligned}.

Rearrange this equation to find an expression for k, the spring constant:

\begin{aligned}k &= \frac{m\, v }{x^{2}}\end{aligned}.

Substitute in the given values m = 1508\; {\rm kg}, v \approx 1.908\; {\rm m\cdot s^{-1}}, and x = 6.87\; {\rm m}:

\begin{aligned}k &= \frac{m\, v }{x^{2}} \\ &\approx \frac{1508\; {\rm kg} \times 1.908\; {\rm m\cdot s^{-1}}}{(6.87\; {\rm m})^{2}} \\ &\approx 1.79 \times 10^{5}\; {\rm kg \cdot m \cdot s^{-2} \cdot m^{-3}}\\ &\approx 1.79 \times 10^{5}\; {\rm N \cdot m^{-1}}\end{aligned}

8 0
2 years ago
What is the first step in allopatric speciation?
fomenos

Answer:

The word Allopatric is derived from the Greek words allos = other and Patris = fatherland. Allopatric specification happens when the biological population of the same specie got separated from one another such that it stopped their gene flow or gene migration. It happens due to the geographical changes such as the movement of continents, changes in glaciers, mountains or oceans and it could be happened because of human activities like agriculture and industrial development etc.  The first step of Allopatric specification is Geographical isolation. The species got separated and the long distances makes them unable to interbreed.

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