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3 years ago

PLZ HELP DUE TONIGHT

Physics

1 answer:

Svetach [21]3 years ago

3 0

By using google, i have found this resource to hep you solve your question.

https://www.siyavula.com/read/science/grade-10/sound/10-sound-03

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Lighting a match results in light and heat. Since energy was not created by the match, what transformation took place?

Yuki888 [10]

Chemical energy stored in the compounds caked on the head of the match, and later the
chemical energy stored in the wood, was released in the form of heat and light when the
chemical compounds got hot enough.

3 0

3 years ago

I’ll give brainliest!! please help and answer correctly! plsss answer quick

Rashid [163]

Answer: The motion of the object will remain the same

Explanation:

6 0

3 years ago

4. The Mariana trench is in the Pacific Ocean and has a depth of approximately 11,000 m. The density of seawater is approximatel

Alex73 [517]

Explanation:

It is known that relation between pressure and density is as follows.

P = $\rho gh$

where, P = pressure

$\rho$ = density

g = acceleration due to gravity

h = height

Putting the given values into the above formula as follows.

P = $\rho gh$

= $1025 \times 9.8 \times 11000$

= 110495000 Pa

Now, relation between pressure and force is as follows.

P = $\frac{F}{A}$

or, F = PA

F = $110495000 \times \pi \times (0.1)^{2}$

= $3.47 \times 10^{6} N$

Thus, we can conclude that a force of $3.47 \times 10^{6} N$ can be experienced at such depth.

3 0

2 years ago

two-point charges are 10.0 cm apart and have charges of 2.0 uc and -2.0uc respectively What is the magnitude of the electrical f

Scorpion4ik [409]

Answer:

Electric field due to two charges is given as

$E = 1.44 \times 10^7 N/C$

Explanation:

As we know that two charges are opposite in nature

So the electric field at the mid point of two charges will add together

so the net field is given as

$E = 2\frac{kq}{r^2}$

now we have

$q = 2\times 10^{-6} C$

$r = 5 cm = 0.05 m$

now we have

$E = 2\frac{(9\times 10^9)(2\times 10^{-6})}{0.05^2}$

$E = 1.44 \times 10^7 N/C$

6 0

2 years ago

A spring of spring constant k is attached to a support at the bottom of a ramp that makes an angle θ with the horizontal. A bloc

Nikitich [7]

Answer:

$x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}$

Explanation:

From the law of conservation of energy

Energy lost by the spring, W=Kinetic energy gained, KE+Potential energy gained, PE+Work done by friction, Fr

$0.5kd^{2}=0.5mv^{2}+mgLsin\theta+\mu_{k}mgcos\theta x$

$x(mgsin\theta+\mu_{k}mgcos\theta)=0.5kd^{2}$

$x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}$

The required distance from A to B is $x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}$

5 0

3 years ago