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3 years ago

5

How can speed be defined

2 answers:

Anna11 [10]3 years ago

6 0

Explanation:

C) distance / time

.......

Tomtit [17]3 years ago

4 0

Answer:

Speed can be defined in (C) distance / time.

I hope this helped at all.

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The hottest climates on Earth are located near the Equator because this region

Ber [7]

You should select Choice-4 .

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3 years ago

____________occurs when an object travels in a curved path.

labwork [276]

Answer:

Acceleration

Explanation:

can you mark me brainlies

So, if an object travels in a curved path, it changes velocity, and, thus, accelerates. This acceleration must be tied to a force. ... Therefore, whenever an object travels in a curved path, there must be an unbalanced force acting upon it. It is important to understand that all this may occur without a change in speed.t

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3 years ago

WOULUJUTUL RECIPECUIUS.

3241004551 [841]

The force between the two objects is 19.73 nN.

<u>Explanation: </u>

Any force acting between two objects tends to be directly proportional to the product of their masses and inversely proportional to the square of the distance between the two objects. And this kind of attraction force between two objects is termed as gravitational force.

So if we consider $M_{1}$ and $M_{2}$ as the masses of both objects and let d be the distance of separation of two objects. Then the force between the two objects can be determined as below:

$\text {Gravitational force}=\frac{G \times M_{1} \times M_{2}}{d^{2}}$

As gravitational constant $G=6.67 \times 10^{-11} \mathrm{m}^{3} \mathrm{kg}^{-1} \mathrm{s}^{-2}$ , $M_{1}$ = 20 kg and $M_{2}$ = 100 kg, while d = 2.6 m, then

$\text {Gravitational force}=\frac{6.67 \times 10^{-11} \times 20 \times 100}{(2.6)^{2}}=\frac{6.67 \times 20 \times 10^{-9}}{6.76}$

Thus, we get finally,

$\text {Gravitational force}=19.73 \times 10^{-9} \mathrm{N}$

As we know, nano denoted by letter 'n' equals to $10^{-9}$

So the force acting between two objects is 19.73 nN.

7 0

3 years ago

Coherent light of frequency 6.37×1014 Hz passes through two thin slits and falls on a screen 88.0 cm away. You observe that the

IgorC [24]

Answer:

The distance between the two slits is 40.11 μm.

Explanation:

Given that,

Frequency $f= 6.37\times10^{14}\ Hz$

Distance of the screen l = 88.0 cm

Position of the third order y =3.10 cm

We need to calculate the wavelength

Using formula of wavelength

$\lambda=\dfrac{c}{f}$

where, c = speed of light

f = frequency

Put the value into the formula

$\lambda=\dfrac{3\times10^{8}}{6.37\times10^{14}}$

$\lambda=471\ nm$

We need to calculate the distance between the two slits

$m\times \lambda=d\sin\theta$

$d =\dfrac{m\times\lambda}{\sin\theta}$

Where, m = number of fringe

d = distance between the two slits

Here, $\sin\theta =\dfrac{y}{l}$

Put the value into the formula

$d=\dfrac{3\times471\times10^{-9}\times88.0\times10^{-2}}{3.10\times10^{-2}}$

$d=40.11\times10^{-6}\ m$

$d = 40.11\ \mu m$

Hence, The distance between the two slits is 40.11 μm.

7 0

3 years ago

A 27 kg bear slides, from rest, 14 m down a lodgepole pine tree, moving with a speed of 6.1 m/s just before hitting the ground.

Nadusha1986 [10]

<h2>Thus the force of friction is 235 N</h2>

Explanation:

When the bear was at the height of 14 m . Its potential energy = m g h

here m is the mass of bear , g is acceleration due to gravity and h is the height .

Thus P.E = 27 x 10 x 14 = 3780 J

The K.E of the bear just before hitting = $\frac{1}{2}$ m v²

= $\frac{1}{2}$ x 27 x ( 6.1 )² = 490 J

The force of friction f = P.E - K.E = 3290 J

Because the work done = Force x Distance

Thus frictional force = $\frac{3290}{14}$ = 235 N

3 0

4 years ago