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BabaBlast [244]
3 years ago
11

A heavy bank-vault door is opened by the application of a force of 300 N directed perpendicular to the plane of the door at a di

stance of 0.80 m from the hinges. What is the torque?
300 Nm

240 Nm

375 Nm

120 Nm
Physics
1 answer:
kondaur [170]3 years ago
5 0
The answer is 240 Nm.
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Explain why wine glasses with long stems have broad base​
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Answer:

The best glasses have a wider bowl than rim to allow for proper swirling. The swirl releases volatile aroma compounds and creates a vortex in the center of the glass towards which these compounds are drawn

Explanation:\\\

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3 years ago
Select the correct location on the image.
slega [8]

Answer:

orange line

Explanation:

it is the only straight line that makes sense. curved lines imply velocity changed. and the completly flat green line implies the vehicle didn't move since its position value stayed the same. must be the downward pointing orange line

8 0
3 years ago
a ball of mass 100g moving at a velocity of 100m/s collides with another ball of mass 400g moving at 50m/s in same direction, if
klio [65]

Answer:

Velocity of the two balls after collision: 60\; \rm m \cdot s^{-1}.

100\; \rm J of kinetic energy would be lost.

Explanation:

<h3>Velocity</h3>

Because the question asked about energy, convert all units to standard units to keep the calculation simple:

  • Mass of the first ball: 100\; \rm g = 0.1\; \rm kg.
  • Mass of the second ball: 400\; \rm g = 0.4 \; \rm kg.

The two balls stick to each other after the collision. In other words, this collision is a perfectly inelastic collision. Kinetic energy will not be conserved. The velocity of the two balls after the collision can only be found using the conservation of momentum.

Assume that the system of the two balls is isolated. Thus, the sum of the momentum of the two balls will stay the same before and after the collision.

The momentum of an object of mass m and velocity v is: p = m \cdot v.

Momentum of the two balls before collision:

  • First ball: p = m \cdot v = 0.1\; \rm kg \times 100\; \rm m \cdot s^{-1} = 10\; \rm kg \cdot m \cdot s^{-1}.
  • Second ball: p = m \cdot v = 0.4\; \rm kg \times 50\; \rm m \cdot s^{-1} = 20\; \rm kg \cdot m \cdot s^{-1}.
  • Sum: 10\; \rm kg \cdot m \cdot s^{-1} + 20 \; \rm kg \cdot m \cdot s^{-1} = 30 \; \rm kg \cdot m \cdot s^{-1} given that the two balls are moving in the same direction.

Based on the assumptions, the sum of the momentum of the two balls after collision should also be 30\; \rm kg \cdot m \cdot s^{-1}. The mass of the two balls, combined, is 0.1\; \rm kg + 0.4\; \rm kg = 0.5\; \rm kg. Let the velocity of the two balls after the collision v\; \rm m \cdot s^{-1}. (There's only one velocity because the collision had sticked the two balls to each other.)

  • Momentum after the collision from p = m \cdot v: (0.5\, v)\; \rm kg \cdot m \cdot s^{-1.
  • Momentum after the collision from the conservation of momentum: 30\; \rm kg \cdot m \cdot s^{-1}.

These two values are supposed to describe the same quantity: the sum of the momentum of the two balls after the collision. They should be equal to each other. That gives the equation about v:

0.5\, v = 30.

v = 60.

In other words, the velocity of the two balls right after the collision should be 60\; \rm m \cdot s^{-1}.

<h3>Kinetic Energy</h3>

The kinetic energy of an object of mass m and velocity v is \displaystyle \frac{1}{2}\, m \cdot v^{2}.

Kinetic energy before the collision:

  • First ball: \displaystyle \frac{1}{2} \, m \cdot v^2 = \frac{1}{2}\times 0.1\; \rm kg \times \left(100\; \rm m \cdot s^{-1}\right)^{2} = 500\; \rm J.
  • Second ball: \displaystyle \frac{1}{2} \, m \cdot v^2 = \frac{1}{2}\times 0.4\; \rm kg \times \left(50\; \rm m \cdot s^{-1}\right)^{2} = 500\; \rm J.
  • Sum: 500\; \rm J + 500\; \rm J = 1000\; \rm J.

The two balls stick to each other after the collision. Therefore, consider them as a single object when calculating the sum of their kinetic energies.

  • Mass of the two balls, combined: 0.5\; \rm kg.
  • Velocity of the two balls right after the collision: 60\; \rm m\cdot s^{-1}.

Sum of the kinetic energies of the two balls right after the collision:

\displaystyle \frac{1}{2} \, m \cdot v^{2} = \frac{1}{2}\times 0.5\; \rm kg \times \left(60\; \rm m \cdot s^{-1}\right)^2 = 900\; \rm J.

Therefore, 1000\; \rm J - 900\; \rm J = 100\; \rm J of kinetic energy would be lost during this collision.

7 0
3 years ago
It is found experimentally that the electric field in a certain region of Earth's atmosphere is directed vertically down. At an
Lemur [1.5K]

Answer:

q=1.7346×10⁻⁶C

Explanation:

Since the electric field is perpendicular to the bottom and top of the cube,the total flux is equals the flux over the top of surface plus the flex over the lower surface

Ф(total)=Ф₃₀₀+Ф₂₃₀

But the flux is given by  Ф=E.A=EACos(θ) where θ is the angle between Area vector and electric field

So

Ф(total)=E₃₀₀A Cos(180)+E₂₃₀ACos(0)

Ф(total)=A(E₃₀₀ - E₂₃₀)

The total flux is given by Gauss Law as:

Ф(total)=q/ε₀

q=ε₀Ф(total)

q=ε₀(A(E₃₀₀ - E₂₃₀))

Substitute the given values

q=(8.85×10⁻¹²){(70²)(100 - 60)}

q=1.7346×10⁻⁶C

7 0
3 years ago
5. A wire, resistance 23.6 carries a current. 5.9 A. What is the
posledela

answer : p.d = 23.6 x 5.9 =139.14

Explanation:

p.d = I x r

3 0
3 years ago
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