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2 years ago

(I will give brainliest whoever helps me !!)

Physics

1 answer:

lorasvet [3.4K]2 years ago

3 0

C. Forces have mass and take up space

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How high would a projectile go if it was launched from ground level with an initial speed of 26 m/s at an angle of 30 degrees ab

tigry1 [53]

Answer:

Vy = 26 m/s sin 30 = 13 m/s vertical speed

t = Vy / a = 13 m/s / 9.80 m/s^2 = 1.33 sec time to reach Vy = 0

H = Vy t + 1/2 g t^2

H = 13 m/s * 1.33 sec - 1.33^2 * 9.8 / 2 m = 8.62 m

4 0

10 months ago

O que é cena fone de luz na visão da fisica

yulyashka [42]

Me don’t speak spanish

4 0

3 years ago

Brut [27]

Answer:

$\boxed{\sf Work \ done = 4 \ J}$

Given:

Force = 8 N

Distance covered by the body = 50 cm = 0.5 m

Explanation:

Work Done = Force × Distance covered by the body

= 8 × 0.5

= 4 J

6 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

On Earth, a scale shows that you weigh 585 N. What would the scale read on the Moon (g = 1.60 m/s2)? N

Serggg [28]

Answer:

95.51 N

Explanation:

First, find the mass in kg:

Fg = 585 N

Fg = m*g

585 N = m*9.8 m/s^2

Then, to find your weight (Fg) on the moon, you use the same equation of

Fg(moon) = m*g, except this time g = 1.60 m/s^2

Fg(moon) = 59.69 kg * 1.60 m/s^2

Fg(moon) = 95.51 N

Hope this helps!! :)

8 0

2 years ago