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Ghella [55]
2 years ago
10

Y’all wann trade in rob.lox I have a violet Valkyrie

Physics
2 answers:
Advocard [28]2 years ago
8 0

Answer:

No no no no no no no no no no no no no

romanna [79]2 years ago
8 0
Pay me, I want 600k by 2:00
You might be interested in
46 points :)
IgorC [24]

Answer:

Its is dividing by 2

Explanation:

It starts with 100 them it goes to 50, 25, 12.5 so its a cycle of dividing by 2

3 0
3 years ago
Read 2 more answers
a 230 kg roller coaster reaches the top of the steepest hill with a speed of 6.2 km/h. It then descends the hill, which is at an
igor_vitrenko [27]

Answer: 81.619 kJ

Explanation:

Given

Mass of roller coaster is m=230\ kg

It reaches the steepest hill with speed of u=6.2\ km/h\ or \ 1.72\ m/s

Hill to bottom is 51 m long with inclination of 45^{\circ}

Height of the hill is h=51\sin 45^{\circ}=36.06\ m

Conserving energy to get kinetic energy at bottom

Energy at top=Energy at bottom

\Rightarrow K_t+U_t=K_b+U_b\\\Rightarrow \dfrac{1}{2}mu^2+mgh=K_b+0\\\\\Rightarrow K_b=0.5\times 230\times 1.72^2+230\times 9.8\times 36.06\\\Rightarrow K_b=340.216+81,279.24\\\Rightarrow K_b=81,619.456\ J\\\Rightarrow K_b=81.619\ kJ

8 0
2 years ago
A 160.-kilogram space vehicle is traveling along a straight line at a constant speed of 800. Meters per second. The magnitude of
natulia [17]

Answer:

Zero

Explanation:

As force acting on the body is equal to the product of mass and acceleration.

Acceleration is equal to rate of change in velocity.

Here velocity is constant so acceleration is zero.

It means the net force acting on the vehicle is zero.

4 0
3 years ago
Read 2 more answers
A ball is dropped from rest at the top of a 6.10 m
natita [175]

Answer:

n = 5 approx

Explanation:

If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back

\frac{v_1}{v} = e ( coefficient of restitution ) = \frac{1}{\sqrt{10} }

and

\frac{v_1}{v} = \sqrt{\frac{h_1}{6.1} }

h₁ is height up-to which the ball bounces back after first bounce.

From the two equations we can write that

e = \sqrt{\frac{h_1}{6.1} }

e = \sqrt{\frac{h_2}{h_1} }

So on

e^n = \sqrt{\frac{h_1}{6.1} }\times \sqrt{\frac{h_2}{h_1} }\times... \sqrt{\frac{h_n}{h_{n-1} }

(\frac{1}{\sqrt{10} })^n=\frac{2.38}{6.1}= .00396

Taking log on both sides

- n / 2 = log .00396

n / 2 = 2.4

n = 5 approx

3 0
3 years ago
Two identical light springs with spring constant k3 are now individually hung vertically from the ceiling and attached at each e
anyanavicka [17]

Answer:

 Keq = 2k₃

Explanation:

We can solve this exercise using Newton's second one

                F = m a

Where F is the eleatic force of the spring F = - k x

Since we have two springs, they are parallel or they are stretched the same distance by the object and the response force Fe is the same for the spring age due to having the same displacement

          F + F = m a

         k₃ x + k₃ x = m a

         a = 2k₃  x / m

To find the effective force constant, suppose we change this spring to what creates the cuddly displacement

       Keq = 2k₃

6 0
3 years ago
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