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oksano4ka [1.4K]
3 years ago
12

Can you help me on this? it is a phyisical science question.

Physics
1 answer:
LenKa [72]3 years ago
8 0

Answer:

Electron

Explanation:

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PLZZZZ HELP QUICK . 30 points. Emma weighs 560 N, she has decided to stay in shape so she is working out every morning, This mor
Phantasy [73]

Force equals mass time acceleration. Weight is a force and it can replace force in the equation. The acceleration would be gravity, which is an acceleration.

1.)

Fw (weight) = m (mass) · g (gravity, 9.8 m/s²)

Fw = m * 9.81 m/s²

560N = m · 9.81 m/s²

m ≈ 57.08 kg

2.)

d = 350 meters

t = 65 seconds

velocity = d/t

velocity = 350 meters / 65 seconds

velocity ≈ 5.38 meters/sec

3.)

Force = 35N

Distance = 2 meters

Work = Force · Distance

Work = 35N · 2 meters

Work = 70 J

3 0
3 years ago
A 1.5 kg bird is gliding at a height of 12 m with a speed of 3.8m/s. What is the kinetic energy of the bird, to the nearest joul
Georgia [21]

A 1.5 kg bird is gliding at a height of 12 m with a speed of 3.8m/s. The kinetic energy of the bird is 10.83 joules.

Explanation:

Kinetic energy can be defined as,The kinetic energy (KE) of an object is the energy that the object possesses due to its motion.

The Kinetic energy can be calculated by using formula,

Kinetic Energy: KE = 1/2 (mv 2)

Where, m = Mass, v = Velocity.

Here in this case the bird mass is 1.5kg and is gliding with velocity 3.8m/s

hence, KE= 1/2*(1.5)×(3.8)^2

                   =0.5×1.5×3.8×3.8

                   =10.83Joules

                   

                   

5 0
3 years ago
How do physicists define velocity?
Natalija [7]

Answer:

Velocity, quantity that designates how fast and in what direction a point is moving.

Explanation:

7 0
3 years ago
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Look at the distance-time graph for a remote-controlled car. What is the speed of the car?
Dafna1 [17]
Speed = distance/time

Speed = 50/5
Speed = 10m/s
7 0
3 years ago
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A lighthouse is located on a small island, 3 km away from the nearest point on a straight shoreline, and its light makes four re
lbvjy [14]

Answer:

The beam of light is moving at the peed of:

\frac{dy}{dt} = \frac{80\pi}{3} km/min

Given:

Distance from the isalnd, d = 3 km

No. of revolutions per minute, n = 4

Solution:

Angular velocity, \omega = \frac{d\theta'}{dt} = 2\pi n = 2\pi \times 4 = 8\pi    (1)

Now, in the right angle in the given fig.:

tan\theta' = \frac{y}{3}

Now, differentiating both the sides w.r.t t:

\frac{dtan\theta'}{dt} = \frac{dy}{3dt}

Applying chain rule:

\frac{dtan\theta'}{d\theta'}.\frac{d\theta'}{dt} = \frac{dy}{3dt}

sec^{2}\theta'\frac{d\theta'}{dt} = \frac{dy}{3dt} = (1 + tan^{2}\theta')\frac{d\theta'}{dt}

Now, using tan\theta = \frac{1}{m} and y = 1 in the above eqn, we get:

(1 + (\frac{1}{3})^{2})\frac{d\theta'}{dt} = \frac{dy}{3dt}

Also, using eqn (1),

8\pi\frac{10}{9})\theta' = \frac{dy}{3dt}

\frac{dy}{dt} = \frac{80\pi}{3}

7 0
3 years ago
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