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aleksklad [387]
3 years ago
10

Can someone help me please

Physics
1 answer:
TEA [102]3 years ago
8 0

Answer:

D

Explanation:

Because it is impossible for it to show the real depth of the ocean and how deep it is

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A scientist examines a large pot of boiling water and a small cup of boiling water. The scientist determines that the large pot
saul85 [17]

Answer:

d. The large pot of water and small cup of water have the same temperature, but the large pot of water has higher thermal energy.

Explanation:

Temperature is a measure of the average kinetic energy of individual molecules. While internal energy refers to the total kinetic energy of the molecules within the object. Since in this case we have the same amount of average kinetic energy, then the large pot of water and small cup of water have the same temperature. While the large pot of water has higher thermal energy, since has more water particles than the small cup.

7 0
3 years ago
Read 2 more answers
6X-6=9<br><br> Solve for X<br><br> Round to TWO decimal places
Brums [2.3K]

Answer:

X=2.50

Explanation:

6x-6=9

6x= 9+6

X=15/6

X= 2.50

8 0
3 years ago
A 0.50 kg toy is attached to the end of a 1.0 m very light string. The toy is whirled in a horizontal circular path on a frictio
xenn [34]

Answer:

The maximum speed will be 26.475 m/sec

Explanation:

We have given mass of the toy m = 0.50 kg

radius of the light string r = 1 m

Tension on the string T = 350 N

We have to find the maximum speed without breaking the string  

For without breaking the string tension must be equal to the centripetal force

So T=\frac{mv^2}{r}

So 350=\frac{0.5\times v^2}{1}

v^2=700

v = 26.475 m /sec

So the maximum speed will be 26.475 m/sec

6 0
3 years ago
Please help, what's the efficiency of the two pulleys​
lutik1710 [3]

Answer:20cm

Explanation:

5 0
3 years ago
A 0.540-kg bucket rests on a scale. Into this bucket you pour sand at the constant rate of 56.0 g/s. If the sand lands in the bu
romanna [79]

Answer:

a) 12.8212 N

b) 12.642 N

Explanation:

Mass of bucket = m = 0.54 kg

Rate of filling with sand  = 56.0 g/ sec = 0.056 kg/s

Speed of sand = 3.2 m/s

g= 9.8 m/sec2

<u>Condition (a);</u>

Mass of sand = Ms = 0.75 kg

So total mass becomes = bucket mass + sand mass = 0.54 +0.75=1.29 kg

== > total weight = 1.29 × 9.8 = 12.642 N

Now impact of sand = rate of filling × velocity = 0.056 × 3.2 =  0.1792 kg. m /sec2=0.1792 N

Scale reading is sum of impact of sand and weight force ;

i-e

scale reading = 12.642 N+0.1792 N = 12.8212 N

<u>Codition (b);</u>

bucket mass + sand mass = 0.54 +0.75=1.29 kg

==>weight = mg = 1.29 × 9.8 = 12.642 N (readily calculated above as well)

6 0
3 years ago
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