3 years ago

2 When a cube of hot metal is placed in a beaker of cold water, the temperature of the water -

Physics

1 answer:

jek_recluse [69]3 years ago

4 0

The answer to this is B

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A carriage of 20 kg is pulled with a force of 35 N. How far the carriage will go

Gennadij [26K]

Answer:

2.71 m

Explanation:

Force is the product of mass and acceleration

F=m*a

Work done is the product of force and distance

Work done=F*d

In this case;

F= 35 N

Work done = 95 J

95 =35 * d

95 /35 = d

2.71 m= d

6 0

3 years ago

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

A cannon at rest fires a 32.5 kg cannonball forward at 388 m/s. After firing, the cannon recoils at 7.42 m/s. What is the mass o

Bond [772]

Answer:

1700 kg

Explanation:

Let’s use conservation of momentum

32.5 * 388 = 7.42 * mc

mc = 1699.46

mc = 1700 kg

3 0

2 years ago

How much force is needed to accelerate a vehicle with a mass of 1000 kg at a rate<br> of 5 m/s2?

Mashcka [7]

The force needed to accelerate a vehicle with a mass of 1000kg at a rate of 5m/s2 would be 5000

8 0

2 years ago

A machine, modeled as a simple spring-mass system, oscillates in simple harmonic motion. Its acceleration is measured to have an

ANTONII [103]

$a=5000\dfrac{mm}{s}=5\dfrac{m}{s}$

$f=10{Hz}\Longrightarrow t=\dfrac{1}{10}s$

$a_{max}=\dfrac{50\frac{m}{s}}{\frac{s}{10}}\cdot\dfrac{1}{\sqrt{2}}=\dfrac{50\dfrac{m}{s^2}}{\sqrt{2}}\approx\boxed{35.4\dfrac{m}{s^2}}$

Hope this helps.

r3t40

5 0

2 years ago