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kap26 [50]
3 years ago
13

A ball is thrown straight down with a speed of Va2.00 from a building 40.00 meters high. How much S time passes before the ball

reaches the ground? a.) -3.07s b.) - 2.46 s c.) =2.66s d.) =9.89s e.) None of the above.A ball is thrown straight down with a speed of Va2.00 from a building 40.00 meters high. How much S time passes before the ball reaches the ground? a.) -3.07s b.) - 2.46 s c.) =2.66s d.) =9.89s e.) None of the above.
Physics
1 answer:
-Dominant- [34]3 years ago
6 0

Answer:

Option C is the correct answer.

Explanation:

Considering vertical motion of ball:-

Initial velocity, u =  2 m/s

Acceleration , a = 9.81 m/s²

Displacement, s = 40 m

We have equation of motion s= ut + 0.5 at²

Substituting

   s= ut + 0.5 at²

    40 = 2 x t + 0.5 x 9.81 x t²

   4.9t²  + 2t - 40 = 0

   t = 2.66 s   or t = -3.06 s

So, time is 2.66 s.

Option C is the correct answer.

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How is newton's first law used in basketball?
arlik [135]

Newton's first law of motion states that an object at rest tends to stay at rest, while an object in motion tends to stay in motion unless an external force acts upon it. This law appears in basketball when the player is shooting the ball. When the player is holding the ball, the ball is at rest but when a player shoots the ball, they use force to throw the ball in the hoop.

4 0
3 years ago
a solid metal sphere of radius 3.00m carries a total charge of -5.50. what is the magnitude of the electric field at a distance
aivan3 [116]

Answer:

(a) Electric field at 0.250 m is zero.

(b)  Electric field at 2.90 m is zero.

(c) Electric field at 3.10 m is - 5.15 x 10³ V/m.

(d) Electric field at 8.00 m is - 0.77 x 10³ V/m.

Explanation:

Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.

Electric field inside and outside the metal sphere is :

E = 0 for r ≤ R ( inside )

  = \frac{KQ}{r^{2} } for r > R ( outside )

Here K is electric constant and r is the distance from the center of the metal sphere.

(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.

(b)  Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.

(c) Electric field at 3.10 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{3.10^{2} }

E = - 5.15 x 10³ V/m

(d) Electric field at 8.00 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{8^{2} }

E = - 0.77 x 10³ V/m

8 0
3 years ago
In a physics lab, a 0.500-kg cart (Cart A) moving with a speed of 129 cm/s encounters a magnetic collision with a 1.50-kg cart (
omeli [17]

Answer:

58 cm/s

Explanation:

0.5×129=0.5×(-45)+1.5×V

V=58

7 0
2 years ago
Un móvil se desplaza con una rapidez inicial de 100 km/h, se le aplican los frenos con una
KIM [24]

Answer:

in English please I am quite puzzled

4 0
3 years ago
In a race, Usain Bolt accelerates at
jeka94

Answer:

65.87 s

Explanation:

For the first time,

Applying

v² = u²+2as.............. Equation 1

Where v = final velocity, u = initial velocity, a = acceleration, s = distance

From the question,

Given:  u = 0 m/s (from rest), a = 1.99 m/s², s = 60 m

Substitute these values into equation 1

v² = 0²+2(1.99)(60)

v² = 238.8

v = √238.8

v = 15.45 m/s

Therefore, time taken for the first 60 m is

t = (v-u)/a............ Equation 2

t = (15.45-0)/1.99

t = 7.77 s

For the final 40 meter,

t = (v-u)/a

Given: v = 0 m/s(decelerates), u = 15.45 m/s, a = -0.266 m/s²

Substitute into the equation above

t = (0-15.45)/-0.266

t = 58.1 seconds

Hence total time taken to cover the distance

T = 7.77+58.1

T = 65.87 s

3 0
3 years ago
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