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bija089 [108]
3 years ago
5

Almost everything in the universe orbits around a central object. The planets orbit the sun and moons orbit planets. The force M

OST responsible for these orbits is
Physics
2 answers:
vovangra [49]3 years ago
5 0
Gravity. The reason the planets are stuck in the sun's orbit is that smaller things get pulled to larger things.
DanielleElmas [232]3 years ago
3 0
Gravity is why orbits are happening.
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How is a coil of current carrying wire similar to a bar magnet
zalisa [80]

Answer:

 When an electric current flows, the shape of the magnetic field is very similar to the field of a bar magnet

Explanation:

5 0
3 years ago
A student constructed a series circuit consisting of a 12.0-volt battery, a 10.0-ohm lamp, and a
Stels [109]
The power dissipated across a component can be calculated through the formula P=I^2xR

Substituting the values in we get P=(0.5)^2x10=2.5W
4 0
3 years ago
A car travels in a straight line for 5 h at a constant speed of 72 km/h. What is it’s acceleration
Luda [366]

Answer:

a \approx \: 0.001 \: m {s}^{ - 2}

Explanation:

Given:

initial \:  velocity \:  (u) = 0 \\  \\ Final  \: Velocity \:  (v) = 72 km /h   \\  \\ Time \:  (t) = 5 \:  hours \\  \\ Acceleration \:  (a) =?  \\  \\  \because \: a =  \frac{v - u}{t}  \\  \\  \therefore \: a =  \frac{72 - 0}{5}  \\  \\ a =  \frac{72}{5}  \\  \\ a = 14.5 \: km {h}^{ - 2}  \\  \\ a =  \frac{14.5 \times 1000}{3600\times 3600} \: m {s}^{ - 2}   \\  \\ a = 0.00111882716 \\  \\ a \approx \: 0.001 \: m {s}^{ - 2}

4 0
3 years ago
1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n
sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

(1).\: x =v_0t

(2).\: y= 15m-\dfrac{1}{2}gt^2

where v_0 is the initial velocity.

(a).

When the projectile hits the 50m mark, y=0; therefore,

0=15-\dfrac{1}{2}gt^2

solving for t we get:

t= 1.75s.

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

50m = v_0(1.75s)

which gives

\boxed{v_0 = 28.58m/s.}

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

v_x = 28.58m/s.

the vertical component of the velocity is

v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.

which gives a speed v of

v = \sqrt{v_x^2+v_y^2}

\boxed{v =33.3m/s.}

4 0
2 years ago
You and a friend are playing with a bowling ball to demonstrate some ideas of Rotational Physics. First, though, you want to cal
RideAnS [48]

Answer:

K_{total} = 19.4 J

Explanation:

The total kinetic energy that is formed by the linear part and the rotational part is requested

         K_{total} = K_{traslation}  + K_{rotation}

let's look for each energy

linear

        K_{traslation} = ½ m v²

rotation

        K_{rotation} = ½ I w²

the moment of inertia of a solid sphere is

       I = 2/5 m r²

we substitute

       K_{total} = ½ mv² + ½ I w²

           

angular and linear velocity are related

           v = w r

we substitute

           K_{total} = ½ m w² r² + ½ (2/5 m r²) w²

           K_{total} = m w² r² (½ + 1/5)

           K_{total} = \frac{7}{10} m w² r²

let's calculate

           K_{total} = \frac{7}{10}   6.40 16.0² 0.130²

           K_{total} = 19.4 J

6 0
2 years ago
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