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serious [3.7K]
2 years ago
13

Joey throws a football 30 meters with a force of 10 N. How much work does Joey do?

Physics
1 answer:
stich3 [128]2 years ago
8 0

Answer:

300 Joules

Explanation:

Just use the Work formula:

W = F . D

W = 10 . 30

W = 300 Joules

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A satellite dish is in the shape of a parabolic surface. Signals coming from a satellite strike the surface of the dish and are
LenKa [72]

Answer: 3.125 ft

Explanation:

If this dish has the form of a concave upward parabola and its vertex p is at the origin, its corresponding equation is:

x^{2}=4py

Where:

x is the radius, which can be found by dividing the diameter d=10 ft by half. Hence x=\frac{d}{2}=\frac{10 ft}{2}=5 ft

y=2 ft is the depth

p is the vertex of the parabola, where its base is

Finding p:

p=\frac{x^{2}}{4y}

p=\frac{(5 ft)^{2}}{4(2 ft)}

Finally:

p=3.125 ft This is where the the receiver should be placed

8 0
3 years ago
What volume of .2500 m cobalt iii chloride is required to react completely with 25 ml of .0315 m calcium hydroxide?
Lynna [10]
<span>f 0.0315 M calcium hydroxide?<span>Co2(SO4)3 + 3Ca(OH)2---> 2Co(OH)3+ 3CaSO4 Co Sulfate and Ca hydroxide react

could you mark as brailiest</span></span>
8 0
3 years ago
In the experiment, “Rolling Along”, which ball had the greater mass?
olga2289 [7]

Answer:

b

Explanation:

4 0
2 years ago
A stone dropped in a pond sends out a circular ripple whose radius increases at a constant rate of 4 ft/sec. After 12 seconds, h
Natali5045456 [20]

Answer:

After 12 seconds, the area enclosed by the ripple will be increasing rapidly at the rate of 1206.528 ft²/sec

Explanation:

Area of a circle = πr²

where;

r is the circle radius

Differentiate the area with respect to time.

\frac{dA}{dt} = 2\pi r\frac{dr}{dt}

dr/dt = 4 ft/sec

after 12 seconds, the radius becomes = \frac{dr}{dt} X 12 = 4 \frac{ft}{sec}  X 12 sec = 48 ft

To obtain how rapidly is the area enclosed by the ripple increasing after 12 seconds, we calculate dA/dt

\frac{dA}{dt} = 2\pi r\frac{dr}{dt}

\frac{dA}{dt} = 2\pi (48)(4)

    dA/dt = 1206.528 ft²/sec

Therefore, after 12 seconds, the area enclosed by the ripple will be increasing rapidly at the rate of 1206.528 ft²/sec

7 0
3 years ago
Calculate the potential energy of a 5 kg object sitting at the top of a 2 meter ramp.
trapecia [35]
The formula used to find potential energy is <em>P.E. = M * G * H</em> (P.E. is potential energy, M is mass, G is gravitational pull, and H is height). So the answer to your question is <em>5 * 9.8 * 2</em>, which equals 98.
5 0
3 years ago
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