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Naddika [18.5K]
3 years ago
5

A sample of a solid nonmetal is compared to a sample of a solid metal. Both samples have the same volume. What is the most likel

y difference between the two samples?
Physics
2 answers:
KATRIN_1 [288]3 years ago
7 0
The weight could be different, metals have a higher mass than nonmetals, so when occupying the same amount of space, the weight of the metal is far more.
Elis [28]3 years ago
6 0
The answer below is correct- the biggest difference would be weight
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Mr Johnson launches an arrow horizontally at a rate of 40m/s off of a 78.4 m cliff towards the south, how much time does it take
vazorg [7]

Answer:c

Explanation:

3 0
3 years ago
A 75.0 kg diver falls from rest into a swimming pool from a height of 5.10 m. It takes 1.34 s for the diver to stop after enteri
vichka [17]

Answer:

559.5 N

Explanation:

Applying,

v² = u²+2gs............. Equation 1

Where v = final velocity,

From the question,

Given: s = 5.10 m, u = 0 m/s ( from rest)

Constant: 9.8 m/s²

Therefore,

v² = 0²+2×9.8×5.1

v² = 99.96

v = √(99.96)

v = 9.99 m/s

As the diver eneters the water,

u = 9.99 m/s, v = 0 m/s

Given: t = 1.34 s

Apply

a = (v-u)/t

a = 9.99/1.34

a = -7.46 m/s²

F = ma.............. Equation 2

Where F = force, m = mass

Given: m = 75 kg, a = -7.46 m/s²,

F = 75(-7.46)

F = -559.5 N

Hence the average force exerted on the diver is 559.5 N

7 0
3 years ago
A system of inertia 0.47 kg consists of a spring gun attached to a cart and a projectile. The system is at rest on a horizontal
Vlad [161]

Answer:

v'=0.83m/a and v=10.2m/s

Explanation:

The information that we have is:

m_1=0.050kg\\m_2=0.47kg\\\theta = 40\\h=2.2m

The maximum height of the projectile is given by the equation

h=\frac{v^2sin^2\theta}{2g}

So, rearrange for the velocity,

v=\sqrt{\frac{2gh}{sin^2\theta}}\\v=\sqrt{\frac{2*9.8*2.2}{sin^2(40)}}\\v=10.2m/s

Apply the conservation of momentum,

mvcos(\theta)=m'v'

Then rearrange the recoil speed,

v'=\frac{mvcos\theta}{m_2}\\v'=\frac{0.05*10.2*cos40}{0.47}\\v'=0.83m/s\\

4 0
3 years ago
A solid cylinder with a mass of 2.46 kg and a radius of 0.049 m starts from rest at a height of 4.80 m and rolls down a 24.3 ◦ s
irina1246 [14]

Answer:

v_{f}\approx 2.097\,\frac{m}{s}

Explanation:

Let assume that the solid cylinder rolls down a frictionless incline. The translational speed can be found by using the Principle of Energy Conservation and the Work-Energy Theorem:

m_{cyl}\cdot g\cdot y_{o} = \frac{1}{2}\cdot m_{cyl} \left( 1+ \frac{1}{R}\right)\cdot v_{f}^{2}

g\cdot y_{o} = \frac{1}{2}\cdot\left( 1+ \frac{1}{R}\right)\cdot v_{f}^{2}

The translational speed is:

v_{f} = \sqrt{\frac{2\cdot g\cdot y_{o}}{\left(1 + \frac{1}{R}  \right)}}

v_{f} = \sqrt{\frac{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (4.80\,m)}{\left(1 + \frac{1}{0.049\,m}  \right)} }

v_{f}\approx 2.097\,\frac{m}{s}

7 0
3 years ago
Tommy was walking at a rate of 4 miles hour at noon and at 12:30 pm he was walking at a brisk rate of 6 miles hour . Two hours l
maks197457 [2]

Answer:

B) Tommy had a positive acceleration between noon and 12:30 pm.

Explanation:

Acceleration is defined as the rate of change of velocity:

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time

In the problem,

- At noon, Tommy is walking at a velocity of 4 mi/h

- At 12.30 pm, Tommy is walking at a velocity of 6 mi/h

- A time of half an hour (0.5 h) passed between the two moments

So Tommy's acceleration is

a=\frac{6 mi/h-4 mi/h}{0.5 h}=4 mi/h^2

and the acceleration is positive, since the velocity has increased.

3 0
3 years ago
Read 2 more answers
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