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lbvjy [14]
3 years ago
6

1. Which of the following is closest to having a mass of 1 kilogram?

Physics
1 answer:
Bingel [31]3 years ago
8 0

Answer:

C. One liter of water

Explanation:

One liter of water is closest to 1 kilogram. The other options don't make sense because a dime cannot be close to a mass of 1 kilogram, neither an average size banana or an average person

<em>PLEASE DO</em><em> </em><em>MARK ME</em><em> </em><em>AS BRAINLIEST</em><em> </em><em>IF MY</em><em> </em><em>ANSWER IS</em><em> </em><em>HELPFUL</em><em> </em><em>;</em><em>)</em><em> </em>

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A 720 g softball is traveling at 15.0 m/s when caught. If the force of the glove on the ball is 520 N, what is the time it takes
Sholpan [36]

Answer:

The time it takes the ball to stop is 0.021 s.

Explanation:

Given;

mass of the softball, m = 720 g = 0.72 kg

velocity of the ball, v = 15.0 m/s

applied force, F = 520 N

Apply Newton's second law of motion, to determine the time it takes the ball to stop;

F = ma = \frac{mv}{t} \\\\t = \frac{mv}{F} \\\\t = \frac{0.72 \ \times \ 15}{520} \\\\t = 0.021 \ s \\

Therefore, the time it takes the ball to stop is 0.021 s.

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3 years ago
Any four function of health post​
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Answer:

1. be the first line of treatment for minor health conditions

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4.  provide basic health services to people who live in rural areas.

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2 years ago
How can the atomic number of nitrogen be determined?
Doss [256]

Answer:

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4 0
2 years ago
A boat of mass 250 kg is coasting, with its engine in neutral, through the water at speed 1.00 m/s when it starts to rain with i
Vera_Pavlovna [14]

Answer:

Explanation:

1. We use the conservation of momentum for before the raining and after. And also we take into account that in 0.5h the accumulated water is

100kg/h*0.5h = 50kg

p_{b}=p_{a}\\M_{b}v_{b}=(M_{b}+m_{r})v_{a}\\v_{a}=\frac{M_{b}v_{b}}{M_{b}+m_{r}}=\frac{250kg*1\frac{m}{s}}{250kg+50kg}=0.83\frac{m}{s}

2. the momentum does not conserve because the drag force of water makes that the boat loses velocity

3. If we assume that the force of the boat before the raining is

F=ma=m\frac{v-v_{0}}{t-t_{0}}=250kg\frac{1m}{s^{2}}=250N

where we have assumed that the acceleration of the boat is 1m/s{2} just before the rain starts

And if we take the net force as

F_{net}=M_{b}a_{net}=F-F_{d}=250N-bv^{2}\\F_{net}=250N-0.5N\frac{s^{2}}{m^{2}}(1\frac{m}{s})^{2}=249.5N\\a_{net}=\frac{249.5N}{M_{b}}=\frac{249.5N}{250kg}=0.99\frac{m}{s^{2}}

where we take v=1m/s because we are taking into account tha velocity just after the rain stars.

I hope this is useful for you

regards

5 0
3 years ago
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