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natta225 [31]
3 years ago
15

Which statement is true about the atom?

Physics
1 answer:
slavikrds [6]3 years ago
7 0

Answer:

Its d

atome contain

negative electrons,

positive protons and uncharged neutrons.

Explanation:

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Visible light is a range of _____ energy EM waves in the electromagnetic spectrum that the human eye can see. A. zero B. high C.
patriot [66]
The answer is a or d

7 0
3 years ago
Read 2 more answers
A 56.0 kg box hangs from a rope. what is the tension in the rope if: the box has vy = 5.30 m/s and is speeding up at 5.10 m/s2 ?
Minchanka [31]

T = tension force in the rope in upward direction

m = mass of the box attached at end of rope = 56 kg

W = weight of the box in downward direction due to gravity

a = acceleration of the box in upward direction = 5.10 m/s²

weight of the box is given as

W = mg

inserting the values

W = (56) (9.8)

W = 548.8 N


force equation for the motion of the box is given as

T - W = ma

inserting the values

T - 548.8 = (56) (5.10)

T = 834.4 N

5 0
3 years ago
A cooling fan is turned off when it is running at 850 rev/min. It turns 1500 revolutions before it comes to a stop. (a) What was
8_murik_8 [283]

Answer

given,

cooling fan revolution = 850 rev/min

fan turns before revolution = 1500 revolutions

\omega = 850 \dfrac{2\pi}{60}

\omega = 89\ rad/s

θ = 1500 revolution

θ = 1500 x 2 x π

θ = 9424.78 rad

a) using equation of rotation

ω² = ω₀² + 2 α θ

ω = 0 because body comes to rest

0 = 89² + 2 x α x 9424.78

α = -0.42 rad/s²

b) time take for the fan to stop

ω = ω₀ + α t

0 = 89 - 0.42 t

t = \dfrac{89}{0.42}

t = 211.9 s

5 0
3 years ago
I HAVE 5 MINUTES!!!!!! A block oscillating on the end of a spring moves from its position of maximum spring stretch to maximum s
Verdich [7]

Therefore, if the block moves from its position of maximum spring stretch to maximum spring compression in 0.25 s, the time required for a full cycle is twice as much; T = 0.5 s.

4 0
3 years ago
For a given velocity of projection in a projectile motion, the maximum horizontal distance is possible only at ө = 45°. Substant
myrzilka [38]

First let us imagine the projectile launched at initial velocity V and at angle θ relative to the horizontal. (ignore wind resistance)

Vertical component y:

The initial vertical velocity is given as Vsinθ
The moment the projectile reaches the maximum height of h, the vertical velocity will be 0, therefore the time t taken to attain this maximum height is:

h = Vsinθ - gt
0 = Vsinθ - gt
t = (Vsinθ)/g

where g is  acceleration due to gravity

Horizontal component x:
The initial horizontal velocity is given as Vcosθ. However unlike the vertical component, this horizontal velocity remains constant because this is unaffected by gravity. The time to travel the horizontal distance D is twice the value of t times the horizontal velocity.
D = Vcosθ*[(2Vsinθ)/g] 
D = (2V²sinθ cosθ)/g 
 D = (V²sin2θ)/g

In order for D (horizontal distance) to be maximum, dD/dθ = 0
That is,

2V^2 cos2θ / g = 0
And since 2V^2/g must not be equal to zero, therefore cos(2θ) = 0
This is true when 2θ = π/2  or  θ = π/4


Therefore it is now<span> shown that the maximum horizontal travelled is attained when the launch angle is π/4 radians, or 45°.</span>

6 0
3 years ago
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