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3 years ago

A swimmer does 3560 j of work in 55 s what is the swimmers power output?

Physics

1 answer:

ankoles [38]3 years ago

8 0

Explanation:

Here it is given 

Work (W) = 3560 J

Time (t) = 55 sec

power (P) = ?

We know we have the formula 

 $p = \frac{w}{t}$ 

P = 3560/55

P = 64.73 watt

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Convert 1 x 10-3 nm to m. given: 1 m = 1,000,000,000 nm

vampirchik [111]

Answer:

l m = 1000000000 nm

? = 0.001

= 0

5 0

3 years ago

If vector A =i+2j-k and vec A cross vec B =3i-j+5k. find vec B

postnew [5]

Let B = a i + b j + c k. Then the cross product of A = i + 2j - k with B is

A × B = ( i + 2j - k ) × ( a i + b j + c k )

A × B = a ( i × i ) + 2a ( j × i ) - a ( k × i )

… … … + b ( i × j ) + 2b ( j × j ) - b ( k × j )

… … … + c ( i × k ) + 2c ( j × k ) - c ( k × k )

A × B = 0 - 2a k - a j

… … … + b k + 0 + b i

… … … - c j + 2c i - 0

A × B = (b + 2c) i + (-a - c) j + (b - 2a) k

So we have

3 i - j + 5 k = (b + 2c) i + (c - a) j + (b - 2a) k

which gives us the system of equations,

{ b + 2c = 3

{ -a - c = -1

{ -2a + b = 5

Solve for a, b, and c.

• Eliminate c from the first two equations:

(b + 2c) + 2 (-a - c) = 3 + 2 (-1)

-2a + b = 1

But -2a + b = 5, and 5 ≠ 1, so there is no such vector B that satisfies the cross product!

8 0

3 years ago

At high speeds, a particular automobile is capable of an acceleration of about 0.540 m/s^2. At this rate, how long (in seconds)

Mama L [17]

Answer:

t = 6.68 seconds

Explanation:

The acceleration of the automobile, $a=0.54\ m/s^2$

Initial speed of the automobile, u = 91 km/hr = 25.27 m/s

Final speed of the automobile, v = 104 km/hr = 28.88 m/s

Let t is the time taken to accelerate from u to v. It can be calculated as the following formula as :

$t=\dfrac{v-u}{a}$

$t=\dfrac{28.88-25.27}{0.54}$

t = 6.68 seconds

So, the time taken by the automobile to accelerate from u to v is 6.68 seconds. Hence, this is the required solution.

5 0

3 years ago

A garden hose having with an internal diameter of 1.1 cm is connected to a (stationary) lawn sprinkler that consists merely of a

trapecia [35]

Answer:

Water leaves the sprinkler at a speed of 2.322 m/sec

Explanation:

We have given internal diameter of the garden $d_1=1.1cm$

Speed of water in the hose is $v_1=0.95m/sec$

Number of holes n = 22

Diameter of each holes $d_2=15cm$

According to continuity equation $A_1v_1=A_2v_2$

$d_1^2\times v_1=22\times d_2^2v_2$

$1.1^2\times 0.95=22\times 0.15^2\times v_2$

$v_2=2.322m/sec$

So water leaves the sprinkler at a speed of 2.322 m/sec

6 0

3 years ago

When a rubber band is pulled back on your finger but not yet let go how is that potential energy?

loris [4]

It is potential energy because the band is not in movement, th band has the potential to move.

4 0

3 years ago