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a_sh-v [17]
3 years ago
6

A ball is thrown upward. After reaching a maximum height, it continues falling back toward Earth. On it way down, the ball is ca

ught at the same height at which it was thrown upward. t hA y tB hB tA b b b b b b b b b b b b b b b b b b b b b b b b v0 O B A Neglecting air resistance, what is its speed when caught?
1. The same as its initial speed
2. Less than its initial speed
3. More than its initial speed
4. Not enough information is given.
Physics
2 answers:
Svet_ta [14]3 years ago
7 0

Answer:

1. The same as its initial speed

Explanation:

As we know that there is no effect of air resistance

so when ball comes to its initial position we can say that there is no effect of air drag and its displacement is also zero

so we will have

W = F.d

W = 0

So work done by the gravity during its whole path must be zero here

so we will have

W_{net} = K_f - K_i

so here we will have

0 = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_f^2

v_f = v_i

so correct answer will be

1. The same as its initial speed

jeka57 [31]3 years ago
4 0

Answer: same as esiential speed

Explanation:

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Two cars travel westward along a straight highway, one at a constant velocity of 85 km/h, and the other at a constant velocity o
spayn [35]

To solve letter a:

d1 = 85t1 = 16 km, 
85t1 = 16, 
t1 = 16 / 85 = 0.1882 h = 11.29 min. 

d2 = 115t2 = 16 km, 
115t2 = 16, 
t2 = 16 / 115 = 0.139 h = 8.35 min. 

t1 - t2 = 11.29 - 8.35 = 2.94 min. 
Car #2 arrives 2.94 minutes sooner.

To solve letter b:

15 min = 1/4 h = 0.25 h. 
d1 = d2, 
115t = 85(t + 0.25), 
115t = 85t + 21.25, 
115t - 85t = 21.25, 
30t = 21.25, 
t = 21.25 / 30 = 0.71 h, 

d = 115 * 0.71 = 81.65 km.

8 0
3 years ago
A 5kg wheel rolls off a flat roof of a 50 m tall building at 12m/s.
Olegator [25]

Explanation:

a) Given in the y direction (taking down to be positive):

Δy = 50 m

v₀ = 0 m/s

a = 10 m/s²

Find: t

Δy = v₀ t + ½ at²

50 m = (0 m/s) t + ½ (10 m/s²) t²

t = 3.2 s

b) Given in the x direction:

v₀ = 12 m/s

a = 0 m/s²

t = 3.2 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (12 m/s) (3.2 s) + ½ (0 m/s²) (3.2 s)²

Δx = 38 m

6 0
3 years ago
The leaves of a tree lose water to the atmosphere via the process of transpiration. A particular tree loses water at the rate of
Gnoma [55]

Answer:

The speed of the sap flowing in the vessel is 1.90 mm/s

Explanation:

Given:

The rate of water loss, Q = 3 × 10 ⁻⁸ m³/s

Number of vessels contained, n = 2000

Diameter of the vessel, D = 100 Mu m

thus, the radius of the vessel, r = 50 × 10⁻⁶ m

Now, the rate of flow is given as:

Q = AV    .............(1)

where, A is the area of the cross-section

V is the velocity

Total area, A = n × (πr²)

substituting the values in the equation (1), we get

3 × 10 ⁻⁸ m³/s = [2000 × (π × (50 × 10⁻⁶)²)] × V

or

V = 1.909 × 10⁻³ m/s or 1.90 mm/s

Hence, the speed of the sap flowing in the vessel is 1.90 mm/s

7 0
3 years ago
Una barra de aluminio que esta a 78 GRADOS CENTIGRADOS entra en contacto con una barra de cobre de la misma longitud y área que
stiks02 [169]

Answer:

Al llegar a su equilibrio térmico ambas barran tendrán una temperatura de 53 grados centígrados.

Explanation:

Dado que una barra de aluminio que está a 78 grados centígrados entra en contacto con una barra de cobre de la misma longitud y área que esta a 28 grados centígrados, y posteriormente se lleva acabo la transferencia de energía entre ambas barras llegando a su equilibrio térmico, para determinar la temperatura a la que ambas barras llegarán se debe realizar el siguiente cálculo:

(78 + 28) / 2 = X

106 / 2 = X

53 = X

Por lo tanto, al llegar a su equilibrio térmico ambas barran tendrán una temperatura de 53 grados centígrados.

8 0
3 years ago
A small steel roulette ball rolls around the inside of a 30 cm diameter roulette wheel. It is spun at 150 rpm, but is slows to 6
liraira [26]

Solution :

Given

Diameter of the roulette ball = 30 cm

The speed ball spun at the beginning = 150 rpm

The speed of the ball during a period of 5 seconds = 60 rpm

Therefore, change of speed in 5 seconds = 150 - 60

                                                                      = 90 rpm

Therefore,

90 revolutions in 1 minute

or In 1 minute the ball revolves 90 times

i.e. 1 min = 90 rev

     60 sec = 90 rev

        1 sec = 90/ 60 rec

         5 sec = $\frac{90}{60}\times 5$

                   = 75 rev

Therefore, the ball made 75 revolutions during the 5 seconds.

7 0
3 years ago
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