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2 years ago

Help ill give you brainliest !!!

Physics

1 answer:

puteri [66]2 years ago

5 0

Answer:

1. B

2. A

3. C

4. B

5. A

6. Muscular strength is different than muscular endurance because of the fact that muscular strength is the amount of force that can be exerted in one instance. Muscular endurance is how long that you can exert that force without being completely exhausted.

7. Some benefits to strength training is the increase in muscular endurance. There is also the benefit of better muscular strength.

Explanation:

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Can someone please help

ki77a [65]

Answer:

Acceleration of that planet is 30 $\frac{m}{s^{2} }$ .

Given:

initial speed of hammer = 0 $\frac{m}{s}$

time = 1 s

distance = 15 m

To find:

Acceleration due to gravity = ?

Formula used:

Distance covered by hammer is given by,

s = ut + $\frac{1}{2} a t^{2}$

s = distance

u = initial speed of hammer

t = time taken by hammer to reach ground

a = acceleration

Solution:

Distance covered by hammer is given by,

s = ut + $\frac{1}{2} a t^{2}$

s = distance

u = initial speed of hammer

t = time taken by hammer to reach ground

a = acceleration

u = 0

t = 1 s

s = 15 m

a = g

Thus substituting these value in above equation.

15 = 0 + $\frac{1}{2} g 1^{2}$

g = 15 × 2

g = 30 $\frac{m}{s^{2} }$

Thus, acceleration of that planet is 30 $\frac{m}{s^{2} }$ .

8 0

3 years ago

A body of mass 200gram is executing SHM with amplitude of 20mm. The magnitude of maximum force which acts upon it is 0.6N. Calcu

igor_vitrenko [27]

The value of maximum velocity will be 0.3464 m/s². Energy or work is equal to the product of force and displacement.

<h3>What is velocity?</h3>

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

Mass of the body is,(m= 200g)

Amplitude is,(A =20 mm)

Maximum force is,F= 0.6 N

To find;

Maximum velocity

Energy or work is equal to the product of force and displacement.

$\rm KE = \frac{1}{2} mV_{max}^2 \\\\ \rm F_{max} \times A = \frac{1}{2} mV_{max}^2 \\\\ F_{max}= \frac{1}{2}\frac{ mV_{max}^2}{A} \\\\ 0.6= \frac{1}{2}\times \frac{ 0.2 \times V_{max}^2}{20 \times 10^{-3}} \\\\ V_{max}=0.3464 \ m/sec^2$