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shusha [124]
3 years ago
9

Measurement conversions [metric to metric] 568 cm = m​

Physics
1 answer:
Trava [24]3 years ago
7 0

Answer:

5.68 meters

Explanation:

hope this helps!

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When a golf club hits a 0.0459 kg ball at rest, it exerts a 2380 N force for 0.00100 s. What is the speed of the ball afterwards
aksik [14]

Answer:

51.85m/s

Explanation:

Given parameters:

Mass of ball  = 0.0459kg

Force  = 2380N

Time taken  = 0.001s

Unknown:

Speed of the ball afterwards  = ?

Solution:

To solve this problem, we use Newton's second law of motion:

   F = m x \frac{v - u}{t}  

F is the force

m is the mass

v is the final velocity

u is the initial velocity

t is the time taken

        2380  = 0.0459 x \frac{v- 0}{0.001}  

        0.0459v  = 2.38

                   v = 51.85m/s

8 0
3 years ago
Read 2 more answers
What type of System interact with its environment ​
Oksi-84 [34.3K]

Answer:

System management

Explanation:

8 0
3 years ago
An object with a mass of 6.0 kg accelerates 8.0 m/s^2 when an unknown force is applied to it. What is the amount of the force? R
iren [92.7K]

Answer:

48N

Explanation:

use F=ma, or force is equal to mass multiplied by acceleration.

8 0
3 years ago
Example 3 :
kherson [118]

The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.

<h3>How to determine the friction factor</h3>

Using the formula

μ = viscosity = 0. 06 Pas

d =  diameter = 120mm = 0. 12m

V =  velocity = 1m/s and 3m/s

ρ = density = 0.9

a. Velocity = 1m/s

friction factor = 0. 52 × \frac{0. 06}{0. 12* 1* 0. 9}

friction factor = 0. 52 × \frac{0. 06}{0. 108}

friction factor = 0. 52 × 0. 55

friction factor = 0. 289

b. When V = 3mls

Friction factor = 0. 52 × \frac{0. 06}{0. 12 * 3* 0. 9}

Friction factor = 0. 52 × \frac{0. 06}{0. 324}

Friction factor = 0. 52 × 0. 185

Friction factor = 0.096

Loss When V = 1m/s

Head loss/ length = friction factor × 1/ 2g × velocity^2/ diameter

Head loss = 0. 289 × \frac{1}{2*6. 6743 * 10^-11} × \frac{1^2}{0. 120} × \frac{1}{100}

Head loss =  1. 80 × 10^8

Head loss When V = 3m/s

Head loss = 0. 096 × \frac{1}{1. 334 *10^-10} × \frac{3^2}{0. 120} × \frac{1}{100}

Head loss = 5. 3× 10^8

Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss  when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.

Learn more about friction here:

brainly.com/question/24338873

#SPJ1

4 0
1 year ago
A physical science test book has a mass of 2.2kg. . What is the weight on the earth
dem82 [27]
Weight = mg,          g ≈ 9.8 m/s²

Weight = 2.2 * 9.8 ≈ 21.56 N
8 0
3 years ago
Read 2 more answers
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