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just olya [345]
3 years ago
13

The weight of an object is the force pulling the object:

Physics
1 answer:
oksano4ka [1.4K]3 years ago
6 0
D pulling because of gravity
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An object is thrown straight up into the air with an initial speed of 8 m/s, and reaches a greatest height of 15 m before it fal
Ugo [173]

Answer:

The value is T_t =  2.5659 \  s    

Explanation:

From the we are told that  

         The initial speed of the object is  u =  8 \  m/s

         The greatest height it reached is  h  =  15 \  m

Generally from kinematic equation we have that

      v^2 =  u^2 + 2gH

At maximum height v  =  0 m/s

So

      0^2 =  8^2 + 2 *  - 9.8 *  H

=>    H  =  3.27 \  m

Here H is the height from the initial height to the maximum height

So the initial height is mathematically represented as  

      s =  h - H

=>    s =  15 - 3.27

=>    s =  11.73 \  m

Generally the time taken for the object to reach maximum height is mathematically evaluated using kinematic equation as follows

            v  =  u + (-g) t

At maximum height v  =  0 m/s

           0 = 8 - 9.8t

=>         t = 0.8163 \  s

Generally the time taken for the object to move from the maximum height to the ground is mathematically using kinematic equation as follows

       h  =  ut_1 + \frac{1}{2}  gt_1^2

Here the initial velocity is  0 m/s given that its the velocity at maximum height

Also  g is positive because we are moving in the direction of gravity  

So

       15  =  0* t  +  4.9 t^2

=>      t_1  =  1.7496

Generally the total time taken is mathematically represented as

          T_t =  t + t_1

=>        T_t =  0.8163  +   1.7496

=>        T_t =  2.5659 \  s            

 

6 0
3 years ago
A baseball player was hit by a baseball in the left eye and complains of double vision. What should you do? A. Restrict spinal m
Alex
I think the smartest answer would be C sorry if i’m writing though
4 0
3 years ago
Read 2 more answers
An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b
lys-0071 [83]

Answer:

a)Distance traveled during the first second = 4.905 m.

b)Final velocity at which the object hits the ground = 38.36 m/s

c)Distance traveled during the last second of motion before hitting the ground = 33.45 m

Explanation:

a) We have equation of motion

             S = ut + 0.5at²

     Here u = 0, and a = g

              S = 0.5gt²

    Distance traveled during the first second ( t =1 )

              S = 0.5 x 9.81 x 1² = 4.905 m

   Distance traveled during the first second = 4.905 m.

b)  We have equation of motion

            v² = u² + 2as

      Here u = 0, s= 75 m and a = g

           v² = 0² + 2 x g x 75 = 150 x 9.81

           v = 38.36 m/s

      Final velocity at which the object hits the ground = 38.36 m/s

c) We have S = 0.5gt²

                   75 = 0.5 x 9.81 x t²

                    t = 3.91 s

   We need to find distance traveled last second

   That is

          S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m

   Distance traveled during the last second of motion before hitting the ground = 33.45 m

       

3 0
3 years ago
Cual sera el caudal que lleva un rio cuando se desplaza a 200 litros cada 40 segundos
alisha [4.7K]

Answer:

Q = 5 L/s

Explanation:

To find the flow you use the following formula (para calcular el caudal usted utiliza la siguiente formula):

Q=\frac{V}{t}

V: Volume (volumen) = 200L

t: time (tiempo) = 40 s

you replace the values of the parameters to calculate Q (usted reemplaza los valores de los parámteros V y t para calcular el caudal):

Q=\frac{200L}{40s}=5\frac{L}{s}

Hence, the flow is 5 L/s (por lo tanto, el caudal es de 5L/s)

4 0
3 years ago
Which formula can be used to find the magnitude of the resultant vector? R2 = Rx2 + Ry2 R = Rx + Ry R = Rx(cosθ) R = Rx(sinθ)
12345 [234]
R^2 = rx^2 + ry^2 !!!!!!!!!!
5 0
2 years ago
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