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Cloud [144]
3 years ago
14

6. question is in the picture

Physics
1 answer:
Harlamova29_29 [7]3 years ago
8 0
The answer is c you got to look for answers that make sense
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How do I solve for horizontal force? And how can I find the acceleration to find the force? And how do I find the power during s
Snezhnost [94]

Answer:hhmm c

Explanation:

5 0
3 years ago
Mia plays soccer for her high school team. generally, they start practice by stretching. what kind of stretching should they do
Keith_Richards [23]

Mia plays soccer for her high school team. generally, they start practicing by stretching. The kind of stretching should they do before they are warmed up is Dynamic Stretching

Dynamic Stretching consists of techniques that involve active stretching to reduce force and its negative effects. Muscles are stimulated and made ready for usage during exercise by dynamic stretching.

Dynamic stretching not only stretches the muscles and tissues but also activates and warms up the muscles to get them ready for action.

We require our muscles' flexibility to maintain the range of motion in our joints, so stretching keeps them strong, flexible, and healthy. The muscles shorten and tighten without it. The muscles are then weak and unable to fully expand when you want them to work.

Learn more about dynamic stretching here

brainly.com/question/14635799

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8 0
1 year ago
What do elements in the same group have?
Natalija [7]
The same number of electron shells, the atomic number is given to each individual element, same as the atomic mass, and properties have nothing to do with groups
5 0
3 years ago
Read 2 more answers
A 0.140 kg baseball is thrown horizontally with a velocity of 28.9 m/s. It is struck with a constant horizontal force that lasts
Viefleur [7K]

Answer:

4987N

Explanation:

Step 1:

Data obtained from the question include:

Mass (m) = 0.140 kg

Initial velocity (U) = 28.9 m/s

Time (t) = 1.85 ms = 1.85x10^-3s

Final velocity (V) = 37.0 m/s

Force (F) =?

Step 2:

Determination of the magnitude of the horizontal force applied. This can be obtained by applying the formula:

F = m(V + U) /t

F = 0.140(37+ 28.9) /1.85x10^-3

F = 9.226/1.85x10^-3

F = 4987N

Therefore, the magnitude of the horizontal force applied is 4987N

8 0
3 years ago
Suppose the ends of a 27-m-long steel beam are rigidly clamped at 0°C to prevent expansion. The rail has a cross-sectional area
frozen [14]

Answer:

F = 1.58*10^{11} N

Explanation:

given data:

length of steel beam = 27 m

cross sectional area of rail = 35 cm

\Delta T = 39 Degree celcius

change in length of steel beam is given as

\Delta L = L_O \alpha \Delta T

            = 20*1.1*10^{-5}*39

           =8.58*10^{-3} m

Young's modulus is

Y = \frac{FL}{A\Delta L}

F = \frac{ YA\Delta L}{L}

= \frac{2.0*10^{11}*25*10^{-4}8.58*10^{-3}}{27}

F = 1.58*10^{11} N

5 0
3 years ago
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