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2 years ago

What does ‘centripetal’ mean in physics?

1 answer:

5 0

Learners Definitionof centripetal force. physics : a force that pulls an object moving in a circular path toward the center of its path.

Centrifugal, however, causes objects to move outwardly.

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Someone help please by providing work and answers please :)

cupoosta [38]

There are two ways to solve this. The longer way is to use those equations to calculate numbers for total distance.

The easier way is to find the area under the graph. That's right, AREA UNDER VELOCITY-TIME graph is the TOTAL DISTANCE travelled!

it's a shortcut.

Let's split up the area into a triangle and rectangle:

Triangle = 0.5(4-0)(10-0) = 20 m
Rectangle = (6-4)(10-0) = 20 m

Total distance = 40 m!

6 0

2 years ago

What is the maximum amount of human weight

Lera25 [3.4K]

Answer:

AN average heavy weight is 175 pounds. The heaviest weight ever recorded though was 1,400 pounds, or 635 kilograms.

7 0

2 years ago

Dinah is playing cards with her friends, and it is her turn to deal. A card with a mass of 2.3 g slides 0.35 m along the table b

Inga [223]

Answer:

soln,

=2.3+0.35-0.24

=2.41 answer

4 0

1 year ago

What might happen if you didn’t run a fever when u get sick

ddd [48]

Answer: then you’re not sick d u h

Explanation:

5 0

2 years ago

If it requires 8.0 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be re

jenyasd209 [6]

Answer:

The amount of work done required to stretch spring by additional 4 cm is 64 J.

Explanation:

The energy used for stretching spring is given by the relation :

$E = \frac{1}{2}kx^{2}$ .......(1)

Here k is spring constant and x is the displacement of spring from its equilibrium position.

For stretch spring by 2.0 cm or 0.02 m, we need 8.0 J of energy. Hence, substitute the suitable values in equation (1).

$8 = \frac{1}{2}\timesk\times k \times(0.02)^{2}$

k = 4 x 10⁴ N/m

Energy needed to stretch a spring by 6.0 cm can be determine by the equation (1).

Substitute 0.06 m for x and 4 x 10⁴ N/m for k in equation (1).

$E = \frac{1}{2}\times4\times10^{4}\times (0.06)^{2}$

E = 72 J

But we already have 8.0 J. So, the extra energy needed to stretch spring by additional 4 cm is :

E = ( 72 - 8 ) J = 64 J

7 0

2 years ago