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LuckyWell [14K]
3 years ago
6

a turtle swimming in a straight line toward shore has a speed of 0.50 m/s after 4.0s its speed is 0.80 m/s what is the turtle's

average acceleration ​
Physics
1 answer:
bagirrra123 [75]3 years ago
5 0

Answer:

sdfv

Explanation:

s.dkvn

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The human heart is a powerful and extremely reliable pump. Each day it takes in and discharges about 7500 L of blood. Assume tha
Yuri [45]

Answer:

(a) 1.257 x 10^5 J

(b) 1.456 Watt

Explanation:

Volume of blood, v = 7500 L = 7.5 m^3

Height, h = 1.63 m

density of blood, d = 1.05 x 10^3 kg/m^3

(a) work done = m x g x h

W = v x d x g x h = 7.5 x 1.05 x 1000 x 9.8 x 1.63 = 1.257 x 10^5 J

(b) time = 1 day = 24 x 60 x 60 s = 86400 seconds

Power = Work / time = 1.257 x 10^5 / 86400 = 1.456 Watt

6 0
3 years ago
Frank has a sample of steel that has a mass of 80 grams if the density is 8g/cm3 what is the volume
nikklg [1K]
Density is mass divided by volume. rho=m/v. So, v=m/rho. In frank's case this is 80/8 = 10 cm^3.
7 0
3 years ago
The mass of a basketball is three times greater than the mass of a softball . Compare the momentum’s of a softball and a basketb
Lemur [1.5K]
They are moving the softball of basketball
4 0
3 years ago
4. As Juan is going to take a shower, the soap falls out of the soap dish on to the
LiRa [457]

The coefficient of friction between the soap and the floor is 0.081

If Juan steps on the soap with a force of 493 N, this is her weight, W. This weight also equals the normal reaction on the floor, N.

We know that frictional force F = μN where μ = coefficient of friction between soap and floor.

So, μ = F/N

Since F = 40 N and N = W = 493 N,

μ = F/N

μ = 40 N/493 N

μ = 0.081

So, the coefficient of friction between the soap and the floor is 0.081

Learn more about coefficient of friction here:

brainly.com/question/13923375

5 0
3 years ago
A 5.75 mm high firefly sits on the axis of, and 11.3 cm in front of, the thin lens A, whose focal length is 5.77 cm . Behind len
weeeeeb [17]

Answer

given,

focal length of lens A = 5.77 cm

focal length of lens B= 27.9 cm

flies distance from mirror = 11.3 m

now,

Using lens formula

\dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{q}

\dfrac{1}{5.77} = \dfrac{1}{11.3} + \dfrac{1}{q}

q =11.79 cm

image of lens A is object of lens B

distance of lens = 59.9 - 11.79 = 48.11

now, Again applying lens formula

\dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{q'}

\dfrac{1}{27.9} = \dfrac{1}{48.11} + \dfrac{1}{q'}

q' =66.41 cm

hence, the image distance from the second lens is equal to q' =66.41 cm

6 0
3 years ago
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