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Nadusha1986 [10]

2 years ago

11

PLEASE HELPPP IT DUES IN 30 minutes

1 answer:

Nuetrik [128]2 years ago

6 0

Answer:

140m

70sec I cant show my work but I gave you a push

Explanation:

Brainliest?

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Calculate the reading on voltmeter v²

Yuliya22 [10]

The reading of the voltmeter can be determined by finding the potential difference across the 2Ω resistance by using the value of current in the circuit. V=IR, here V is the potential difference across a resistance R through which a current I is flowing.

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2 years ago

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A hollow sphere of radius 0.200 m, with rotational inertia I = 0.0484 kg·m2 about a line through its center of mass, rolls witho

d1i1m1o1n [39]

Answer:

Part a)

$KE_r = 8 J$

Part b)

v = 3.64 m/s

Part c)

$KE_f = 12.7 J$

Part d)

$v = 2.9 m/s$

Explanation:

As we know that moment of inertia of hollow sphere is given as

$I = \frac{2}{3}mR^2$

here we know that

$I = 0.0484 kg m^2$

R = 0.200 m

now we have

$0.0484 = \frac{2}{3}m(0.200)^2$

$m = 1.815 kg$

now we know that total Kinetic energy is given as

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2$

$20 = \frac{1}{2}(1.815)v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2$

$20 = 1.5125 v^2$

$v = 3.64 m/s$

Part a)

Now initial rotational kinetic energy is given as

$KE_r = \frac{1}{2}I(\frac{v}{R})^2$

$KE_r = \frac{1}{2}(0.0484)(\frac{3.64}{0.200})^2$

$KE_r = 8 J$

Part b)

speed of the sphere is given as

v = 3.64 m/s

Part c)

By energy conservation of the rolling sphere we can say

$mgh = (KE_i) - KE_f$

$1.815(9.8)(0.900sin27.1) = 20- KE_f$

$7.30 = 20 - KE_f$

$KE_f = 12.7 J$

Part d)

Now we know that

$\frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{r})^2 = 12.7$

$\frac{1}{2}(1.815) v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2 = 12.7$

$1.5125 v^2 = 12.7$

$v = 2.9 m/s$

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3 years ago

if you knew the number of valence electrons in a nonmetal atom, how would you determine the valence of the element?

Vanyuwa [196]

It would be funny because . I will not be good

8 0

2 years ago

Are forces of attraction among the particles in all mater? true or false.

Luden [163]

Answer:

the answer is true

Explanation:

3 0

2 years ago

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Hawks and gannets soar above the ground and, when they spot prey, they fold their wings and essentially drop like a stone. They

denis-greek [22]

Answer:

v = 54.2 m / s

Explanation:

Let's use energy conservation for this problem.

Starting point Higher

Em₀ = U = m g h

Final point. Lower

$Em_{f}$ = K = ½ m v²

Em₀ = Em_{f}

m g h = ½ m v²

v² = 2gh

v = √ 2gh

Let's calculate

v = √ (2 9.8 150)

v = 54.2 m / s

3 0

2 years ago