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Allushta [10]
3 years ago
6

Which of the following are equal to -40°F?

Physics
2 answers:
ValentinkaMS [17]3 years ago
8 0

Answer: -40 F and 233 K

Explanation:

Apex

Marianna [84]3 years ago
3 0

I think It may be -40degreeF is equal to -80degreeC.

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Sinkholes and caves are 2 reasons.  hope that helped
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The smallest unit of an element that has all of the properties of the element is a/an
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Answer:B

Explanation: an atom is the smallest particle of an element that can take part in a chemical reaction.

7 0
3 years ago
An 800-N billboard worker stands on a 4.0-m scaffold weighing 500 N and supported by vertical ropes at each end. How far would t
Lynna [10]

Answer:

2.5 m

Explanation:

Weight of billboard worker = 800 N

Number of ropes = 2

Length of scaffold = 4 m

Weight of scaffold = 500 N

Tension in rope = 550 N

The sum of the torques will be

-800(4-x)-500\times 2+550\times 4=0\\\Rightarrow -800(4-x)=500\times 2-550\times 4\\\Rightarrow -800(4-x)=-1200\\\Rightarrow -x=\dfrac{1200}{800}-4\\\Rightarrow -x=-2.5\\\Rightarrow x=2.5\ m

The position of the person will be 2.5 m

7 0
3 years ago
An object with a mass of 2.0 kg is accelerated at 5.0 m/s/s. the net force acting on the mass is
Vitek1552 [10]
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7 0
3 years ago
Read 2 more answers
A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the
Margaret [11]

Hi there!

\boxed{\omega = 0.38 rad/sec}

We can use the conservation of angular momentum to solve.

\large\boxed{L_i = L_f}

Recall the equation for angular momentum:

L = I\omega

We can begin by writing out the scenario as a conservation of angular momentum:

I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)

I_m = moment of inertia of the merry-go-round (kgm²)

\omega_m = angular velocity of merry go round (rad/sec)

\omega_f = final angular velocity of COMBINED objects (rad/sec)

I_b = moment of inertia of boy (kgm²)

\omega_b= angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

I = MR^2

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

\omega = \frac{v}{r}

L_b = MR^2(\frac{v}{R}) = MRv

Plug in the given values:

L_b = (20)(3)(5) = 300 kgm^2/s

Now, we must solve for the boy's moment of inertia:

I = MR^2\\I = 20(3^2) = 180 kgm^2

Use the above equation for conservation of momentum:

600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}

8 0
2 years ago
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