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dalvyx [7]
3 years ago
12

Is diverse families positive or negative

Physics
1 answer:
nataly862011 [7]3 years ago
6 0
Think its Positive 
hope this helpes
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7. A 50 kg pole vaulter running at 10 m/s vaults over the bar. Her speed when she is over the bar is 1.0 m/s. Neglect air resist
luda_lava [24]

Answer:  5.05 m

Explanation:

4 0
3 years ago
Read 2 more answers
When a 25-kg crate is pushed across a frictionless horizontal floor with a force of 200 N, directed 20 below the horizontal, th
Fofino [41]

Answer:

Option E is correct 310N

Explanation:

Given that the force used to push the crate is F = 200N

The force directed 20° below the horizontal

Mass of crate is m = 25kg

Weight of the crate can be determine using

W = mg

g is gravitational constant =9.8m/s²

W = 25×9.8

W = 245 N

Check attachment. For free body diagram and better understanding

Using newton second law along the vertical axis since we want to find the normal force

ΣFy = m•ay

ay = 0, since the body is not moving in the vertical or y direction

N—W—F•Sin20 = 0

N = W+F•Sin20

N = 245+ 200Sin20

N = 245 + 68.4

N = 313.4 N

The normal force is approximately 310 N to the nearest ten

3 0
3 years ago
2. An egg rolls off the kitchen counter with a velocity of 4m/s. If
-Dominant- [34]

Answer:

800m/s

Explanation:

6 0
3 years ago
Q 2 Two mirrors meet at right angles. A ray of light is incident on one at an angle of 30°
serg [7]

a ray of light is incident towards a plane mirror at an angles of 30degrees with the mirror surface. what will be the angles of reflection is 60degree.

3 0
3 years ago
A baseball of mass m = 0.31 kg is spun vertically on a massless string of length L = 0.51m. The string can only support a tensio
natulia [17]

Given data:

* The mass of the baseball is 0.31 kg.

* The length of the string is 0.51 m.

* The maximum tension in the string is 7.5 N.

Solution:

The centripetal force acting on the ball at the top of the loop is,

\begin{gathered} T+mg=\frac{mv^2}{L}_{} \\ v^2=\frac{L(T+mg)}{m} \\ v=\sqrt[]{\frac{L(T+mg)}{m}} \end{gathered}

For the maximum velocity of the ball at the top of the vertical circular motion,

v_{\max }=\sqrt[]{\frac{L(T_{\max }+mg)}{m}}

where g is the acceleration due to gravity,

Substituting the known values,

\begin{gathered} v_{\max }=\sqrt[]{\frac{0.51(7.5_{}+0.31\times9.8)}{0.31}} \\ v_{\max }=\sqrt[]{\frac{0.51(10.538)}{0.31}} \\ v_{\max }=\sqrt[]{17.34} \\ v_{\max }=4.16\text{ m/s} \end{gathered}

Thus, the maximum speed of the ball at the top of the vertical circular motion is 4.16 meters per second.

8 0
1 year ago
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