C.

An object on Earth weighs 150 N. What is its mass?

Alex777 [14]

Answer:

$\tt \: mass \: = \frac{weight}{acceleration \: due \: to \: gravity}$

$\longrightarrow \tt \: \frac{150}{10}$

$\longrightarrow \boxed{ \tt{15 \: kg}}$

Our final answer is 15 kg .

----------- HappY LearninG<3 ------------

3 0

2 years ago

engineers are using computer models to study train collisions design safer train cars. They start by modeling an elastic collisi

Archy [21]

Answer:

The final velocity of the second car is 57 m/s south.

Explanation:

This is an elastic collision between two train cars. In this case, the total kinetic energy between the two bodies will remain the same.

The formula to apply is :

$m_1v_1i +m_2v_2i=m_1v_1f+m_2v_2f$

where ;

$m_1=mass of object 1\\v_1i=initial velocity object1\\m_2=mass of object2\\v_2i=initial velocity object 2\\v_1f=final velocity object 1\\v_2f=final velocity object 2$

Given in the question that;

$m_1=14650kg\\v_1i=18m/s\\m_2=3825kg\\v_2i=11m/s\\v_1f=6m/s\\v_2f=?$

Apply the formula as;

$m_1v_1i +m_2v_2i=m_1v_1f+m_2v_2f$

{14650*18}+{3825*11} = {14650 *6} + {3825 * v₂f}

263700+42075=87900 + 3825v₂f

305775 =87900 + 3825v₂f

305775-87900 = 3825v₂f

217875=3825v₂f

217875/3825 =v₂f

56.96 = v₂f

<u>57 m/s = v₂f { nearest whole number}</u>

4 0

2 years ago

Read 2 more answers

You are standing 2.5m directly in front of one of the two loudspeakers. They are 3.0m apart and both are playing a 686Hz tone in

ahrayia [7]

Answer:

distance from speaker is 17.87 m

Explanation:

given data

distance from loudspeaker = 2.5 m

distance between loudspeaker = 3.0 m

room temperature = 20c

wavelength f = 686Hz

to find out

what distances from the speaker

solution

we know sound velocity c = 331.5 + 0.6 × 20c = 343.5

so wavelength of sound λ = c / f

wavelength = 343.5 / 686 = 0.5 m

when the difference in distance of speaker destructive interference will be

d = λ/2 × (2n-1)

for n = 1, 2 3 4 ..

d = 0.5/2 × (2n-1)

d = 0.250 , 0.75 , 1.25 , 1.750............ for n = 1, 2 3 .............

so

for d = 0.250

side of triangle by hypotenuse of triangle are

$\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x1)$ = 0.250

0.5 x1 = 7.6875

x1 = 15.375 m

for d = 0.75

side of triangle by hypotenuse of triangle are

$\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x2)$ = 0.75

1.5 x2 = 4.6875

x2 = 3.125 m

for d = 1.250

side of triangle by hypotenuse of triangle are

$\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x3)$ = 1.250

2.5 x2 = 1.1875

x3 = 0.475 m

for d = 1.750

x4 will be negative so we stop here

so the distance from speaker here is given below

distance = 2.5 + x

here x = 0.475 , 3.125 and 15.375 so

distance 1 = 2.5 + 0.475 = 2.975 m

distance 2 = 2.5 + 3.125 = 5.625 m

distance 3 = 2.5 + 15.375 = 17.875 m

final distance from speaker is 17.87 m

8 0

3 years ago

I need help, ASAP, I know what thermal energy is, but how does a skater use it when on the track?

Flauer [41]

Answer: Looked it up but

Explanation:

When the skater lands on the track, the vertical component of his kinetic energy is converted to thermal energy. You can do experiments where there is no loss to thermal energy (only PE and KE conversions) by turning friction off and by making sure the skater doesn't leave the track.

4 0

3 years ago

What would the final temperature be if 8.94 X 10 3 joules of heat were added to 454 grams of copper at 23.0 o C?

ozzi

Answer: $74.1^0C$