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kodGreya [7K]
3 years ago
5

I will mark as the brainliest answerplz 8,9,10​

Physics
1 answer:
blagie [28]3 years ago
5 0

Answer:

8.  acceleration = \dfrac{d(velocity)}{d(time)}  = 1 unit .

9. acceleration = \dfrac{d(velocity)}{d(time)}  = -1 unit.

10. acceleration = \dfrac{d(velocity)}{d(time)}  = 0 units.

Explanation:

8. i) acceleration = velocity / time

  ii) In this figure velocity = time

  iii) therefore acceleration = \dfrac{d(velocity)}{d(time)} = 1 unit .

9. i) acceleration = velocity / time

  ii) In this figure 4 = m + 5, therefore m = -1

     therefore velocity = (-0.5 \times time) + 5

  iii) therefore acceleration = \dfrac{d(velocity)}{d(time)}  = -1 units.

10.) velocity is constant at 2

     therefore acceleration = \dfrac{d(velocity)}{d(time)}  = 0 units

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A tiger leaps horizontally out of a tree that is 6.00 m high. If he lands 2.00 m from the base of the tree, calculate his initia
Musya8 [376]

Answer:

The initial speed of the tiger is 1.80 m/s

Explanation:

Hi there!

The equation of the position vector of the tiger is the following:

r = (x0 + v0 · t, y0 + 1/2 · g · t²)

Where:

r = position vector at a time t.

x0 = initial horizontal position.

v0 = initial horizontal velocity.

t = time.

y0 = initial vertical position,

g = acceleration due to gravity.

Let´s place the origin of the frame of reference on the ground at the point where the tree is located so that the initial position vector will be:

r0 = (0.00, 6.00) m

We can use the equation of the vertical component of the position vector to obtain the time it takes the tiger to reach the ground.

y = y0 + 1/2 · g · t²

When the tiger reaches the ground, y = 0:

0 = 6.00 m - 1/2 · 9.81 m/s² · t²

2 · (-6.00 m) / -9.81 m/s² = t²

t = 1.11 s

We know that in 1.11 s the tiger travels 2.00 m in the horizontal direction. Then, using the equation of the horizontal component of the position vector we can find the initial speed:

x = x0 + v0 · t

At t = 1.11 s, x = 2.00 m

x0 = 0

2.00 m = v0 · 1.11 s

2.00 m / 1.11 s = v0

v0 = 1.80 m/s

The initial speed of the tiger is 1.80 m/s

4 0
3 years ago
Yashoda prepares some lime juice on a hot day. She adds 80 g of ice at a temperature of 0°C to 0.32 kg of lime juice. The temper
Vikentia [17]

Answer:

Explanation:

a )

hear energy required to melt 1 g of ice = 340 J ,

hear energy required to melt 80 g of ice = 340 x 80  J = 27220 J .

b ) energy gained by the melted ice ( water at O°C ) = m ct

where m is mass of water , s is specific heat and t is rise in temperature

= 80 x 4.2 x ( 8°C - 0°C)

= 2688 J .

c )

energy lost by lime juice = energy gained by ice and water

= 27220 J + 2688 J .

= 29908 J .

d )

Let specific heat required be S

Heat lost by lime juice = M S T

M is mass of lime juice , S is specific heat , T is decrease in temperature

= 320 g x S x ( 29 - 8 )°C

= 6720 S

For equilibrium

Heat lost = heat gained

6720 S = 29908 J

S = 4.45 J /g °C .

4 0
3 years ago
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