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3 years ago

Help me plzzzz

Physics

2 answers:

Vinil7 [7]3 years ago

6 0

Answer:

C. Mg3N2

How I Know :

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KiRa [710]3 years ago

4 0

Answer:

C. Mg3N2

Explanation:

Magnesium has a positive 2 charge and nitrogen has a negative 3 charge. To completely balance it out, you need three magnesium’s and two nitrogen’s. The final charge for magnesium would be positive 6 and the final charge for nitrogen would be negative 6, this will make it neutral.

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A 49 kg bear slides, from rest, 11 m down a lodgepole pine tree, moving with a speed of 3.3 m/s just before hitting the ground.

hichkok12 [17]

Answer:

a) $\Delta U_g=-5.3kJ$

b) $K=0.27kJ$

c) $F_f=0.45kN$

Explanation:

the gravitational potential energy is given by:

$U_g=m.g.h\\$

$\Delta U_g=m.g.h_f-m.g.h_i\\\Delta U_g=49kg*9.8m/s^2*(0m-11m)\\\Delta U_g=-5.3kJ$

The kinetic energy is given by:

$K=\frac{1}{2}m.v^2\\$

the initial kinetic energy is zero because the motion started from rest, so:

$K=\frac{1}{2}*49kg*(3.3m/s^2)^2\\K=0.27kJ$

applying the conservation of energy theorem:

$U_g-W_f=K_f\\W_f=-(\Delta K+\Delta U)\\W_F=5.3kJ-0.27kJ\\W_F=-5.0kJ$

The work done by the friction force is given by:

$W_f=F_f.h.cos(\theta)\\$

the angle of the force is 180 degrees because it's against the movement:

$F_f=\frac{W_f}{h.cos(\theta)}\\\\F_f=\frac{-5.0kJ}{11m.cos(180^o)}\\\\F_f=0.45kN$

8 0

3 years ago

You pull with a force of 295 N on a rope that is attached to a block of mass 22 kg, and the block slides across the floor at a c

Sergeeva-Olga [200]

Answer:

Fnet = 0

Explanation:

Since the block slides across the floor at constant speed, this means that it's not accelerated.
According Newton's 2nd Law, if the acceleration is zero, the net force on the sliding mass must be zero.
This means that there must be a friction force opposing to the horizontal component of the applied force, equal in magnitude to it:

$F_{appx} = F_{app} * cos \theta = 295 N * cos 35 = 242 N (1)$

In the vertical direction, the block is not accelerated either, so the sum of the normal force and the vertical component of the applied force, must be equal in magnitude to the force of gravity on the block:

$F_{appy} = F_{app} * cos \theta = 295 N * sin 35 = 169 N (2)$

⇒ 169 N + Fn = Fg = 216 N (3)

This means that there must be a normal force equal to the difference between Fappy and Fg, as follows:
Fn = 216 N - 169 N = 47 N (4)

6 0

2 years ago

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QveST [7]

In a moving car the outside looks to be moving. however if viewed from the outside, the car appears to be moving. so motion is relative to the person observing.

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3 years ago

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DerKrebs [107]

Answer:

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