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2 years ago

You are playing a game of soccer. Describe as many actionreaction pairs in this situation as you can think of.

Physics

1 answer:

marissa [1.9K]2 years ago

3 0

Slide tackle, goalkeeping, avoiding getting tackled, controlling the ball in the air, shooting from a hard angle etc

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Can someone please answer this, ill give you brainliest Would be very appreciated.

Evgesh-ka [11]

Answer:

False

Explanation:

Solid water (ice) has a lesser density than liquid water. Hence, the statement is false.

3 0

1 year ago

Read 2 more answers

A merry-go-round is spinning at a rate of 4.04.0 revolutions per minute. Cora is sitting 0.50.5 m from the center of the merry-g

dsp73

Answer:

angular speed of both the children will be same

Explanation:

Rate of revolution of the merry go round is given as

f = 4.04 rev/min

so here we have

$f = \frac{4.04}{60} =0.067 rev/s$

here we know that angular frequency is given as

$\omega = 2\pi f$

$\omega = 2\pi(0.067)$

$\omega = 0.42 rad/s$

now this is the angular speed of the disc and this speed will remain same for all points lying on the disc

Angular speed do not depends on the distance from the center but it will be same for all positions of the disc

7 0

2 years ago

How do scientists know the continents were once closer than they are today? Explain your answer in 2-3 complete sentences.

WITCHER [35]

Answer:

Continental drift describes one of the earliest ways geologists thought continents moved over time. Today, the theory of continental drift has been replaced by the science of plate tectonics.

The theory of continental drift is most associated with the scientist Alfred Wegener. In the early 20th century, Wegener published a paper explaining his theory that the continental landmasses were “drifting” across the Earth, sometimes plowing through oceans and into each other. He called this movement continental drift.

6 0

2 years ago

Besides gravity, what factor keeps the moon and Earth in orbit?

earnstyle [38]

Answer:

interna

Explanation:

please mark as brainllest

8 0

2 years ago

Read 2 more answers

An object weighs 63.8 N in air. When it is suspended from a force scale and completely immersed in water the scale reads 16.8 N.

I am Lyosha [343]

Answer:

The density of this object is approximately $1.36\; {\rm kg \cdot L^{-1}}$ .

The density of the oil in this question is approximately $0.600\; {\rm kg \cdot L^{-1}}$ .

(Assumption: the gravitational field strength is $g =9.806\; {\rm N \cdot kg^{-1}}$ )

Explanation:

When the gravitational field strength is $g$ , the weight $(\text{weight})$ of an object of mass $m$ would be $m\, g$ .

Conversely, if the weight of an object is $(\text{weight})$ in a gravitational field of strength $g$ , the mass $m$ of that object would be $m = (\text{weight}) / g$ .

Assuming that $g =9.806\; {\rm N \cdot kg^{-1}}$ . The mass of this $63.8\; {\rm N}$ -object would be:

$\begin{aligned} \text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{63.8\; {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 6.506\; {\rm kg}\end{aligned}$ .

When an object is immersed in a liquid, the buoyancy force on that object would be equal to the weight of the liquid that was displaced. For instance, since the object in this question was fully immersed in water, the volume of water displaced would be equal to the volume of this object.

When this object was suspended in water, the buoyancy force on this object was $(63.8\; {\rm N} - 16.8\; {\rm N}) = 47.0\; {\rm N}$ . Hence, the weight of water that this object displaced would be $47.0 \; {\rm N}$ .

The mass of water displaced would be:

$\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{47.0\: {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 4.793\; {\rm kg}\end{aligned}$ .

The volume of that much water (which this object had displaced) would be:

$\begin{aligned}\text{volume} &= \frac{\text{mass}}{\text{density}} \\ &\approx \frac{4.793\; {\rm kg}}{1.00\; {\rm kg \cdot L^{-1}}} \\ &\approx 4.793\; {\rm L}\end{aligned}$ .

Since this object was fully immersed in water, the volume of this object would be equal to the volume of water displaced. Hence, the volume of this object is approximately $4.793\; {\rm L}$ .

The mass of this object is $6.50\; {\rm kg}$ . Hence, the density of this object would be:

$\begin{aligned} \text{density} &= \frac{\text{mass}}{\text{volume}} \\ &\approx \frac{6.506\; {\rm kg}}{4.793\; {\rm L}} \\ &\approx 1.36\; {\rm kg \cdot L^{-1}} \end{aligned}$ .

(Rounded to $\text{$3$ sig. fig.}$ )

Similarly, since this object was fully immersed in oil, the volume of oil displaced would be equal to the volume of this object: approximately $4.793\; {\rm L}$ .

The weight of oil displaced would be equal to the magnitude of the buoyancy force: $63.8\; {\rm N} - 35.6\; {\rm N} = 28.2\; {\rm N}$ .

The mass of that much oil would be:

$\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{28.2\: {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 2.876\; {\rm kg}\end{aligned}$ .

Hence, the density of the oil in this question would be:

$\begin{aligned} \text{density} &= \frac{\text{mass}}{\text{volume}} \\ &\approx \frac{2.876\; {\rm kg}}{4.793\; {\rm L}} \\ &\approx 0.600\; {\rm kg \cdot L^{-1}} \end{aligned}$ .

(Rounded to $\text{$3$ sig. fig.}$ )

7 0

1 year ago