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Fiesta28 [93]
3 years ago
14

If two swimmers compete in a race, does the faster swimmer develop more power?

Physics
1 answer:
valkas [14]3 years ago
7 0
Power is equal to energy per unit time. In this case, power is proportional to energy while is inversely proportional to time,on the other hand. Given the two swimmers exerts same amount of energy but the faster swimmer just does things in faster time, then the faster swimmer should develop more power from shorter time
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A ball is thrown with a speed of 20 m/s at an angle of 40o above the horizontal from the top of a 22-m tall building.
MrRissso [65]

Answer:

(a). The distance is  49.79 m.

(b). The speed of the ball is 24.39 m/s.

Explanation:

Given that,

Speed = 20 m/s

Angle = 40°

Height = 22 m

Time = 3.25 sec

(a). We need to calculate the distance

Using formula of distance

d=u\cos\theta\times t

Put the value into the formula

d=20\cos40\times3.25

d=49.79\ m

(b). We need to calculate the horizontal velocity

Using formula of velocity

v_{x}=u\cos\theta

Put the value into the formula

v_{x}=20\times\cos40

v_{x}=15.3\ m/s

We need to calculate the vertical velocity

Using equation of motion

v_{y}=u\sin\theta-gt

Put the value into the formula

v_{y}=20\sin40-9.8\times3.25

v_{y}=-19\ m/s

Negative sign shows the opposite direction.

We need to calculate the speed of ball

Using formula of speed

v=\sqrt{v_{x}^2+v_{y}^2}

v=\sqrt{(15.3)^2+(19)^2}

v=24.39\ m/s

Hence, (a). The distance is  49.79 m.

(b). The speed of the ball is 24.39 m/s.

4 0
3 years ago
A ball is dropped from rest at the top of a 6.10 m
natita [175]

Answer:

n = 5 approx

Explanation:

If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back

\frac{v_1}{v} = e ( coefficient of restitution ) = \frac{1}{\sqrt{10} }

and

\frac{v_1}{v} = \sqrt{\frac{h_1}{6.1} }

h₁ is height up-to which the ball bounces back after first bounce.

From the two equations we can write that

e = \sqrt{\frac{h_1}{6.1} }

e = \sqrt{\frac{h_2}{h_1} }

So on

e^n = \sqrt{\frac{h_1}{6.1} }\times \sqrt{\frac{h_2}{h_1} }\times... \sqrt{\frac{h_n}{h_{n-1} }

(\frac{1}{\sqrt{10} })^n=\frac{2.38}{6.1}= .00396

Taking log on both sides

- n / 2 = log .00396

n / 2 = 2.4

n = 5 approx

3 0
3 years ago
A pencil has a density of 0.875 g/ml. It has a volume of 4.0 ml. Find the mass
loris [4]

Answer:

To find the mass using density and volume we just multiply them against each other which causes ml to cancel and just leaves us with grams which represents how much the item weights.

mass=density*volume

mass=0.875\frac{g}{ml}*4.0\ ml

mass=3.5\ g

Therefore, our final answer is that our pencil weight 3.5 grams

<u><em>Hope this helps!  Let me know if you have any questions</em></u>

3 0
2 years ago
An element has the following natural abundances and isotopic masses: 90.92% abundance with 19.99 amu, 0.26% abundance with 20.99
sashaice [31]

<u>Answer:</u> The average atomic mass of the given element is 20.169 amu.

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of the isotopes each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i     .....(1)

We are given:

  • For isotope 1:

Mass of isotope 1 = 19.99 amu

Percentage abundance of isotope 1 = 90.92 %

Fractional abundance of isotope 1 = 0.9092

  • For isotope 2:

Mass of isotope 2 = 20.99 amu

Percentage abundance of isotope 2 = 0.26%

Fractional abundance of isotope 2 = 0.0026

  • For isotope 3:

Mass of isotope 3 = 21.99 amu

Percentage abundance of isotope 3 = 8.82%

Fractional abundance of isotope 3 = 0.0882  

Putting values in equation 1, we get:

\text{Average atomic mass}=[(19.99\times 0.9092)+(20.99\times 0.0026)+(21.99\times 0.0882)]

\text{Average atomic mass}=20.169amu

Hence, the average atomic mass of the given element is 20.169 amu.

4 0
3 years ago
Two racing boats set out from the same dock and speed away at the same constant speed of 104 km/h for half an hour (0.500 h), th
Zinaida [17]

Answer:

The blue boat traveled 6.1 km farther west than the green boat

The green boat traveled 10.7 km farther south than the blue boat

Explanation:

The equation for linear uniform speed movement is

X(t) = X0 + v * t

Since we have two coordinates (X, Y) we use

X(t) = X0 + vx * t

Y(t) = Y0 + vy * t

The dock will be the origin of coordinates so X0 and Y0 will be zero. The X axis will be pointing west and the Y axis south.

The blue boat moves with a direction 24° south of west, so it will have speeds:

vxb = 104 * cos(24) = 95 km/h

vyb = 104 * sin(24) = 42.3 km/h

And the green boat:

vxg = 104 * cos(37.7) = 82.3 km/h

vyg = 104 * sin(37.7) = 63.6 km/h

After half hour the boats will have arrived at positions

Xb = 95 * 0.5 = 47.5 km

Yb = 42.3 * 0.5 = 21.1 km

And

Xg = 82.3 * 0.5 = 41.4 km

Yg = 63.6 * 0.5 = 31.8 km

The difference in positions of the boats

47.5 - 41.4 = 6.1 km

31.8 - 21.1 = 10.7 km

5 0
3 years ago
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