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4 years ago

HURRY PLEASE ILL MARK YOU AS BRAINLIEST! 30 POINTS!

Physics

2 answers:

saveliy_v [14]4 years ago

6 0

The correct answer is B .

Electrically charged particles

there i gave you another person answer

kap26 [50]4 years ago

5 0

The correct answer is B .

Electrically charged particles.

You might be interested in

A 59.3 kg diver jumps off a board

ser-zykov [4K]

The board is 2.50m high.

Why?

We can calculate how high was the board applying the Law of Conservation of Mechanical Energy. This Law states that the mechanical energy (kinematic and potential) will be conserved during the motion.

It can be described with the following formula:

$E_{M_{1}}=E_{M_{2}}\\\\PE_{1}+KE_{1}=PE_{2}+KE_{2}$

$PE=m*g*h\\KE=\frac{1}{2}m*v^{2}$

At the top of the boar, the kinetic energy is equal to 0.

At the water, the potential energy is equal to 0.

So,

$PE_{1}=KE_{2}\\\\m*g*h=1450J\\\\59.3kg*9.8\frac{m}{s^{2}}*h=1450J\\ \\h=\frac{1450J}{59.3kg*9.8\frac{m}{s^{2}}}=2.50m$

Hence, we have that the board is 2.50m high.

Have a nice day!

6 0

4 years ago

10. A golf ball is propelled with an initial velocity of 60

AlladinOne [14]

Answer:

See the answer below.

Explanation:

To find the horizontal component, we use the cos(θ).

$V_x=60\times cos(37^\circ)\\\\\approx 46\: m/s$

Best Regards!

4 0

3 years ago

Read 2 more answers

Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L

antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

$\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}$

Similarly, for rod 2

$\alpha_2 = \dfrac{\tau_2}{I_2}.$

Now, the moment of inertia for rod 1 is

$I_1 = \dfrac{1}{3}ML^2$ ,

and the torque acting on it is (about the center of mass)

$\tau_1 = Mg\dfrac{L}{2};$

therefore, the angular acceleration of rod 1 is

$\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},$

$\boxed{\alpha_1 = \dfrac{3g}{2L} }$

Now, for rod 2 the moment of inertia is

$I_2 = \dfrac{1}{3}(2M)(2L)^2$

$I_2 = \dfrac{8}{3} ML^2,$

and the torque acting is (about the center of mass)

$\tau _2 = (2M)g \dfrac{(2L)}{2}$

$\tau _2 = 2MgL;$

therefore, the angular acceleration $\alpha_2$ is

$\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.$

$\boxed{\alpha_2 = \dfrac{3g}{4L}}$

We see here that

$\dfrac{3g}{2L} > \dfrac{3g}{4L}$

therefore

$\boxed{\alpha_1 > \alpha_2.}$

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

7 0

3 years ago

For for each situation say how the ideas of force pressure and area can be applied :

fomenos

<h3 /><h3><u><em>Solution-:</em></u></h3><h3><u><em>more force as expansion is much</em></u></h3>

<u><em>also, less force area is much</em></u>

<h3> mark me Brainliest</h3>

4 0

2 years ago

The lightest and heaviest flying birds are the bee hummingbird of Cuba, which weighs about 1.6 grams, and the great bustard of E

VMariaS [17]

Answer:

for the birds to be able to stay vertical in flight without falling down to earth, they must produce a lift that will counteract their weight

for the small bee humming bird,

mass = 1.6 g = 1.6 x $10^{-3}$ kg

weight of the bird under acceleration due to gravity = mg

where g = acceleration due to gravity = 9.81 m/s^2

weight of the bird = 1.6 x $10^{-3}$ x 9.806 = 0.0156 ≅ 0.016 N

for this bird to maintain flight, the least lift upward, it must generate must be equal to its weight downwards, i.e

lift = weight

therefore,

lift = <em>0.016 N</em>

For the bustard of Europe and Asia,

mass = 21 kg

weight of the bird under acceleration due to gravity = mg

weight of the bird = 21 x 9.806 = 205.9 N

lift = weight = <em>205.9 N</em>

<em>lift generated is proportional to the wing surface area according to the lift equation</em>

L = Cs x p x $\frac{v}{2}$ x S

where L = lift

C = lift coefficient

p = density of air

v = relative velocity of bird and air

S = surface are of the wing.

<em>The great bustard will have a proportionally larger wing area to hold its weight in flight</em>

7 0

3 years ago