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kicyunya [14]
3 years ago
14

Why are large astronomical bodies such as planets and stars round

Physics
2 answers:
Tamiku [17]3 years ago
5 0

Answer:

Gravity pulled little bits of space rocks (or gases) with equivalent force in all directions when they formed .

Explanation:

Alborosie3 years ago
4 0
Gravity pulled little bits of space rocks (or gases) with equivalent force in all directions when they formed 
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(1 point) A projectile is fired from ground level with an initial speed of 600 m/sec and an angle of elevation of 30 degrees. Us
konstantin123 [22]

Answer:

(a) The range of the projectile is 31,813.18 m

(b) The maximum height of the projectile is 4,591.84 m

(c) The speed with which the projectile hits the ground is 670.82 m/s.

Explanation:

Given;

initial speed of the projectile, u = 600 m/s

angle of projection, θ = 30⁰

acceleration due to gravity, g = 9.8 m/s²

(a) The range of the projectile in meters;

R = \frac{u^2sin \ 2\theta}{g} \\\\R = \frac{600^2 sin(2\times 30^0)}{9.8} \\\\R = \frac{600^2 sin (60^0)}{9.8} \\\\R = 31,813.18 \ m

(b) The maximum height of the projectile in meters;

H = \frac{u^2 sin^2\theta}{2g} \\\\H = \frac{600^2 (sin \ 30)^2}{2\times 9.8} \\\\H = \frac{600^2 (0.5)^2}{19.6} \\\\H = 4,591.84 \ m

(c) The speed with which the projectile hits the ground is;

v^2 = u^2 + 2gh\\\\v^2 = 600^2 + (2 \times 9.8)(4,591.84)\\\\v^2 = 360,000 + 90,000.064\\\\v = \sqrt{450,000.064} \\\\v = 670.82 \ m/s

5 0
3 years ago
how would the velocity of the book change if the applied force were equal to the sliding friction force
Inessa [10]
Yes, Sliding friction opposes the movement of the book, slowing it down.sliding That's the 'kinetic' kind.. According to Newton's second law, F=ma. That is, the bear's acceleration should be proportional to the total force acting on the bear. As the bear's velocity is constant, its acceleration is zero. Therefore, the total Force acting on the bear is zero. Thus, the friction has to be equal in magnitude and opposite in direction to the bear's weight. As W=mg, we get that its weight is  <span>9.8*400=3,920 Newton. Thus, the friction acting on the bear is 3,920 Newton</span>
3 0
3 years ago
Which of the following is the best definition of lifestyle activity?
Bezzdna [24]
It d bro it’s d bro it’s d
3 0
3 years ago
A 0.520-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of
Veseljchak [2.6K]

Answer:

Explanation:

Given that,

Mass attached to spring

M = 0.52kg

Force constant K = 8N/m

Amplitude A = 11.6 cm

a. Maximum speed?

Angular velocity is calculated using

w = √k/m

w = √8/0.52

w = √15.385

w = 3.922rad/s

Then, the relation ship between angular velocity and linear velocity is given as

v = - w•A

v = - 3.922 × 11.6

v = - 45.5 cm/s

Then, the maximum velocity is

vmax = |v|= 45.5cm/s

b. Acceleration a?

Acceleration can be determine using the formula

a = -w²• A

a = -3.922² × 11.6

a = -178.46 cm/s²

Magnitude of the acceleration is 178.46cm/s²

c. Speed when the object is at 9.6cm from equilibrium position?

Generally,

The position of the object at equilibrium is

x(t) = A•Cos(wt)

x(t) = 11.6 Cos (3.922t)

Then, when x(t) = 9.6cm

9.6 = 11.6 Cos(3.92t)

Cos(3.922t) = 9.6/11.6

Cos(3.922t) = 0.8276

3.922t = ArcCos(0.8276)

Note: the angle is in radiant

3.922t = 0.596

t = 0.596/3.922

t = 0.152 second

Then, v(t) at that time is

v(t) = x'(t) = -11.6×3.92Sin(3.922t)

v(t) = -45.5Sin(3.922t)

Now, when t =0.152

v(t) = -45.5 Sin(3.922×0.152)

v(t) = -45.5Sin(0.596)

v(t) = -25.5 cm/s

Then, it's magnitude is 25.5cm/s

d. Acceleration at same position

t = 0.152s

a(t) = v'(t) = - 45.5×3.922Cos(3.922t)

a(t) = -178.46Cos(3.92t)

a(t) = -178.46 Cos(3.92×0.152)

a(t) = -178.46 Cos(0.596)

a(t) = -147.68 cm/s²

Magnitude of the acceleration is 147.68 cm/s²

5 0
3 years ago
if the roller coaster is 127 meters tall and has 622,300 J of potential energy how much does it weigh/ what is the mass?
Xelga [282]

Answer:

500 kg

Explanation:

using the formula Potential energy (p.e)=mgh where g= acceleration due to gravity whose value is 9.8

so,pe =mgh

622300=m*9.8*127

m=500kg

4 0
3 years ago
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