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AysviL [449]
3 years ago
6

The game was played by swinging the big mallet down hard enough to cause the bell to ring. If it took a 44 newton force to ring

the bell, and George got it to go only 75% of the way up, how many newtons of force did George exert with the mallet on the bell?
A.
44 newtons
B.
66 newtons
C.
33 newtons
D.
11 newtons
Physics
1 answer:
nata0808 [166]3 years ago
3 0
I think C. 33 newtons
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(-5)/3 - 6/(-5)
You can solve it now :)
7 0
3 years ago
This is a problem about a child pushing a stack of two blocks along a horizontal floor. The masses of the blocks, and the coeffi
Rufina [12.5K]

Answer:

 N = 23.4 N

Explanation:

After reading that long sentence, let's solve the question

The contact force is the so-called normal in this case we can find it by writing the translational equilibrium equation for the y axis

            N - w₁ -w₂ =

            N = m₁ g + m₂ g

            N = g (m₁ + m₂)

let's calculate

            N = 9.8 (0.760 + 1.630)

            N = 23.4 N

This is the force of the support of the two blocks on the surface.

7 0
2 years ago
Two birds have the same mass but one has bigger feet. How would pressure that they produce on the ground be different ?
Mama L [17]

Answer:

the pressure per square inch is greater from the smaller feet.

Explanation:

different weight distribution

3 0
3 years ago
The maximum distance which our normal eye can see distinctly is known as _______________
azamat

Answer:

Far point.

Explanation:

The maximum distance up to which the normal eye can see objects distinct and clear is called the far point of the eye. It is infinity for a normal eye.

7 0
2 years ago
Two carts, one of mass 2m and one of mass m, approach each other with the same speed, v. When the carts collide, they hook toget
Nikitich [7]

Answer:

<em>Second option</em>

Explanation:

<u>Linear Momentum</u>

The linear momentum of an object of mass m and speed v is

P=mv

If two or more objects are interacting in the same axis, the total momentum is

P_t=m_1v_1+m_2v_2+...

Where the speeds must be signed according to a fixed reference

The images show a cart of mass 2m moves to the left with speed v since our reference is positive to the right

P_1=-2mv

The second cart of mass m goes to the right at a speed v

P_2=mv

The total momentum before the impact is

P_t=-2mv+mv=-mv

The total momentum after the collision is negative, both carts will join and go to the left side

The first option shows both carts with the same momentum before the collision and therefore, zero momentum after. It's not correct as we have already proven

The third option shows the 2m cart has a positive greater momentum than the other one. We have proven the 2m car has negative momentum. This option is not correct either

The fourth option shows the two carts keep separated after the collision, which contradicts the condition of the question regarding "they hook together".

The second option is the correct one because the mass m_2 has a negative momentum and then the sum of both masses keeps being negative

3 0
3 years ago
Read 2 more answers
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