All categories

3 years ago

What is 1/3 of 300

Physics

2 answers:

USPshnik [31]3 years ago

6 0

Answer:

100

Explanation:

Phantasy [73]3 years ago

6 0

Answer:

100

Explanation:

100 goes into 300 3 times

You might be interested in

Can someone give me the units

Anettt [7]

Answer:

The unit for power is the Watt

8 0

2 years ago

When an object is lifted 12 meters off the ground, it gains a certain amount of potential energy. If the same object is lifted 2

alexandr402 [8]

I think the answer is twice as much

8 0

1 year ago

Find the magnitude of the sum of two vectors; A is 5 km , and B is 7 , when the angle btween them is 120

Dimas [21]

Answer: 6.24 km

Explanation:

Given

The magnitude of the first vector(say) $\left | a\right |=5\ km$

the magnitude of the second vector(say) $\left | b\right |=7\ km$

the angle between them is $120^{\circ}$

The resultant vector magnitude is given by

$\left | \vec{R}\right |=\sqrt{a^2+b^2+2ab\cos \theta}$

$\left | \vec{R}\right |=\sqrt{5^2+7^2+2\times 5\times 7\cdot \cos 120^{\circ}}\\\left | \vec{R}\right |=\sqrt{74-35}=\sqrt{39}\\\left | \vec{R}\right |=6.24\ km$

4 0

2 years ago

A 97 kg man lying on a surface of negligible friction shoves a 62 g stone away from himself, giving it a speed of 2.6 m/s. What

tangare [24]

Answer:

man will move in opposite direction with speed

$v_1 = 1.66 \times 10^{-3} m/s$

Explanation:

As we know that man is lying on the friction-less surface

so here net force along the surface is zero

so if we take man + stone as a system then net change in momentum of this system will become zero

so here we have

$P_i = P_f$

$0 = m_1v_1 + m_2v_2$

here we have

$0 = (97)v_1 + 0.062(2.6)$

$v_1 = -\frac{0.1612}{97}$

$v_1 = -1.66 \times 10^{-3} m/s$

3 0

2 years ago

A particle travels clockwise on a circular path of diameter R, monitored by a sensor on the circle at point P; the other endpo

kotykmax [81]

We make a graphic of this problem to define the angle.

The angle we can calculate through triangle relation, that is,

$sin\theta = \frac{c}{QP}\\sin\theta = \frac{c}{R}\\\theta=sin^{-1}\frac{c}{R}$

With this function we should only calculate the derivate in function of c

$\frac{d\theta}{dc} = \frac{1}{\sqrt{1-\frac{c^2}{R^2}}}(\frac{c}{R})'\\\frac{d\theta}{dc} = \frac{1}{\sqrt{R^2-c^2}}$

That is the rate of change of $\theta$ .

b) At this point we need only make a substitution of 0 for c in the equation previously found.

$\frac{d\theta}{dc}\big|_{c=0} = \frac{1}{\sqrt{R^2-0}}\\\frac{d\theta}{dc}\big|_{c=0} = \frac{1}{R}$

Hence we have finally the rate of change when c=0.

6 0

2 years ago

What is 1/3 of 300​

What is 1/3 of 300