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3 years ago

What units could be used for speed

Physics

2 answers:

Scrat [10]3 years ago

6 0

Answer:

mph (miles per hour) and kph (kilometers per hour)

In-s [12.5K]3 years ago

3 0

Mph (miles per hour) nfoejnfinweifniewqnf

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Lesson 11: electricity & magnetism unit test physical science b unit 5: electricity and magnetism (connexus)

vovikov84 [41]

Electricity and magnetism can be considered as part of the same phenomenon because both are generated by electromagnetic forces.

<h3>What is electricity?</h3>

Electricity is the movement or flow of negatively charged electrons through a suitable conductor where a charge is applied.

Moreover, magnetism is a natural phenomenon caused by the generation/movement of electric charges.

In conclusion, electricity and magnetism can be considered as part of the same phenomenon because both are generated by electromagnetic forces.

Learn more on electricity and magnetism here:

brainly.com/question/25144822

#SPJ1

5 0

2 years ago

Two identical loudspeakers 2.00 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standi

Troyanec [42]

Answer:

The lowest possible frequency of sound is 971.4 Hz.

Explanation:

Given that,

Distance between loudspeakers = 2.00 m

Height = 5.50 m

Sound speed = 340 m/s

We need to calculate the distance

Using Pythagorean theorem

$AC^2=AB^2+BC^2$

$AC^2=2.00^2+5.50^2$

$AC=\sqrt{(2.00^2+5.50^2)}$

$AC=5.85\ m$

We need to calculate the path difference

Using formula of path difference

$\Delta x=AC-BC$

Put the value into the formula

$\Delta x=5.85-5.50$

$\Delta x=0.35\ m$

We need to calculate the lowest possible frequency of sound

Using formula of frequency

$f=\dfrac{nv}{\Delta x}$

Put the value into the formula

$f=\dfrac{1\times340}{0.35}$

$f=971.4\ Hz$

Hence, The lowest possible frequency of sound is 971.4 Hz.

8 0

3 years ago

The law of conservation of momentum states that the total momentum of interacting objects does not change . This means the total

pickupchik [31]

Answer:

The momentum of an object is equal to the product of its mass and its velocity.

Explanation:

Consider an object of mass $m$ travelling at a velocity $\vec{v}$ . The momentum $\vec{p}$ of this object would be:

$\vec{p} = m \cdot \vec{v}$ .

For the law of conservation of momentum, consider two objects: object $\rm a$ and object $\rm b$ . Assume that these two objects collided with each other.

Let $m_{\rm a}$ and $m_{\rm b}$ denote the mass of the two objects.
Let $\vec{v}_{\rm a}(\text{initial})$ and $\vec{v}_{\rm b}(\text{initial})$ denote the velocity of the two object right before the interaction.
Let $\vec{v}_{\rm a}(\text{final})$ and $\vec{v}_{\rm b}(\text{final})$ denote the velocity of the two objects right after the interaction.

The momentum of the two objects right before the collision would be $m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial})$ and $m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial})$ , respectively.
The momentum of the two objects right after the collision would be $m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final})$ and $m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ , respectively.

The sum of the momentum of the two objects would be:

$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial})$ right before the collision, and
$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ right after the collision.

Assume that the system of these two objects is isolated. By the law of conservation of momentum, the sum of the momentum of these two objects should be the same before and after the collision. That is:

$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial}) = m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ .

4 0

4 years ago

When two point charges are a distance d part, the electric force that each one feels from the other has magnitude F. In order to

Serjik [45]

Answer:

E) d/sqrt2

Explanation:

The initial electric force between the two charge is given by:

$F=k\frac{q_1 q_2}{d^2}$

where

k is the Coulomb's constant

q1, q2 are the two charges

d is the separation between the two charges

We can also rewrite it as

$d=\sqrt{k\frac{q_1 q_2}{F}}$

So if we want to make the force F twice as strong,

F' = 2F

the new distance between the charges would be

$d'=\sqrt{k\frac{q_1 q_2}{(2F)}}=\frac{1}{\sqrt{2}}\sqrt{k\frac{q_1 q_2}{(2F)}}=\frac{d}{\sqrt{2}}$

so the correct option is E.

8 0

3 years ago

PLEASE HELP ASAP WILL REWARD BRAINLIEST:))

laiz [17]

Answer:

120s^-1

Explanation:

v=12v

I=10A

and since rate is with time, therefore rate=energy/time.

H=IV

10×12=120/s

therefore the rate is 120s^-1

6 0

2 years ago