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4 years ago

What is the net force acting on a 2kg obgect that accelerates from rest at a rate of 5m/s/s/?

Physics

1 answer:

Alekssandra [29.7K]4 years ago

8 0

Explanation:

mass=2kg

acceleration=5m/s/s

Force=mass×acceleration

Force=2×5

Force=10N

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3 years ago

The Sears Tower is nearly 400 m high. How long would it take a steel ball to reach the ground if dropped on the top? What will b

kipiarov [429]

Answers:

a) 9.035 s

b) -88.543 m/s

Explanation:

The described situation is related to vertical motion (especifically free fall) and the equations that will be useful are:

$y=y_{o}+V_{o}t+\frac{1}{2}gt^{2}$ (1)

$V=V_{o}+gt$ (2)

Where:

$y=0$ is the final height of the steel ball

$y_{o}=400 m$ is the initial height of the steel ball

$V_{o}=0$ is the initial velocity of the steel ball (it was dropped)

$V$ is the final velocity of the steel ball

$t$ is the time it takes to the steel ball to reach the ground

$g=-9.8 m/s^{2}$ is the acceleration due to gravity

<u>Knowing this, let's begin with the answers:</u>

<h2>a) Time it takes the steel ball to reach the ground</h2>

We will use equation (1) with the conditions listed above:

$0=y_{o}+\frac{1}{2}gt^{2}$ (3)

Isolating $t$ :

$t=\sqrt{\frac{-2y_{o}}{g}}$ (4)

$t=\sqrt{\frac{-2(400 m)}{-9.8 m/s^{2}}}$ (5)

$t=9.035 s$ (6)

<h2>b) Final velocity of the steel ball</h2>

We will use equation (2) with the conditions explained above and the calculaated time:

$V=gt$ (7)

$V=(-9.8 m/s^{2})(9.035 s)$ (8)

$V=-88.543 m/s$ (9) The negative sign indicates the direction of the velocity is downwards

3 0

3 years ago

A 75-hp (shaft output) motor that has an efficiency of 91.0 percent is worn out and is to be replaced by a high- efficiency moto

daser333 [38]

Convert the shaft ouput from HP to kW

Shaft output = 75HP = 55.93kW

1st: Finding for the power consumption based on 55.93kW output

Power consumption (Old) = 55.93kW / .910 = 61.46kW

Power consumption (New) = 55.93kW / .954 = 58.63kW

2nd: Total power used in kWh:

Power Used = Power consumption * load factor * Hours:

Power (Old) = 61.46kW * .75 * 4368 = 201343 kWh

Power (New) = 58.63kW * .75 * 4368 = 192072 kWh

Energy saved = 201343 kWh - 192072 kWh = 9,271 kWh

3rd: Calculating for the price:

Price = kW-Hr * $/kWh

Price (Old) = 201343kWh * $0.08/kWh = $16107.44

Price (New) = 192072 kWh * $0.08/kWh = $15365.76

Cost saved = $16107.44 - $15365.76 = $741.68/yr

4th: Setting up the cost equation:

Cost over time, F(t) = Motor_Cost + (Price * Number of Years, t)

Cost (Old) = 5449 + 16107.44*t

Cost (New) = 5520 + 15365.76*t

Equate the two to find for t when they cost equally:

5449 + 16107.44*t = 5520 + 15365.76*t

16107.44*t = 15365.76*t +71

16107.44*t - 15365.76*t = 71

741.68*t = 71

t = 71 / 741.68 = .095 years = 35 days

So the payback period is after 35 days.

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3 years ago