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SpyIntel [72]
3 years ago
5

A tiger leaps horizotally from a 15.5 m high rock with a speed of 7.5 m/s. How

Physics
1 answer:
aniked [119]3 years ago
5 0

The tiger lands 13.4 m from the base of the rock

Explanation:

The motion of the tiger is a projectile motion, therefore it consists of two independent motions:  

- A uniform motion (constant velocity) along the horizontal direction  

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction  

We start by analyzing the vertical motion, to find the time of flight of the tiger. Using the following suvat equation:

s=u t+\frac{1}{2}at^2

where, taking downward as positive direction, we have:

s = 15.5 m is the vertical displacement of the tiger (the height of the rock)

u=0 is the initial vertical velocity

t is the time of the fall

a=g=9.8 m/s^2 is the acceleration of gravity

Solving for t, we find :

t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(15.5)}{9.8}}=1.78 s

Therefore the tiger takes 1.78 s to land on the ground.

Now we analyze the horizontal motion: the tiger moves horizontally with a constant velocity of

v_x = 7.5 m/s

So, the distance it covers in a time t is given by:

d=v_x t

and by substituting t = 1.78 s, we find the horizontal distance covered by the tiger:

d=(7.5)(1.78)=13.4 m

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

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krek1111 [17]

Answer:

Explanation:

We shall apply the formula for velocity in case of elastic collision which is given below

v₁ = (m₁ - m₂)u₁ /  (m₁ +  m₂)  + 2m₂u₂ / (m₁ +  m₂)

m₁ and u₁ is mass and velocity of first object , m₂ and  u₂ is mass and velocity of second object before collision and v₁ is velocity of first velocity after collision.

Here u₁ = 22 cm /s , u₂ = - 14 cm /s . m₁ = 7.7 gm , m₂ = 18 gm

v₁ = ( 7.7 - 18 ) x 22 / ( 7.7 + 18 )  + 2 x 18 x - 14 / ( 7.7 + 18 )

= - 8.817 - 19.6

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3 years ago
What will a spring scale read for the weight of a 75.0-kg woman in an elevator that moves upward with constant speed of 5.8 m/s
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<span>On the scale the only external forces are the man's weight acting downwards and the normal force which the scale exerts back to support his weight. So F = Ma = mg + Fs The normal force Fs (which is actually the reading on the scale) = Ma + Mg But a = 0 So Fs = Mg which is just his weight. Fs = 75 * 9.8 = 735N</span>
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3 years ago
Rays traveling parallel to the principle axis of a concave mirror will reflect out through the mirrors focus?
abruzzese [7]

Answer:

True

Explanation:

When a ray travelling parallel to the principle axis of a concave mirror then the light ray reflect out through the mirrors and passing through the focus.

When a light ray travelling through focus of a concave mirror then after reflection the light ray reflect out through the mirror and  go parallel to principle axis.

Therefore, rays travelling parallel to the principle axis of a  concave mirror will reflect out through the mirrors focus.

It is true.

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3 years ago
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Susan is quite nearsighted; without her glasses, her far point is 34 cm and her near point is 17 cm . Her glasses allow her to v
butalik [34]

Answer:

u=34cm

Explanation:

From the question we are told that:

Far point is V=34 cm

Near point is u=17 cm

Therefore

Focal Length

f=-34cm

Generally the equation for the Lens is mathematically given by

\frac{1}{u}=\frac{1}{f}-\frac{1}{v}

\frac{1}{u}=\frac{1}{-34}-\frac{1}{-17}

u=34cm

5 0
2 years ago
In Challenge Example 11.9 (p. 280), after the explosion, suppose that the m1 fragment shot directly north at 12 m/s and the m3 f
Inga [223]

The question is incomplete. The mass of the object is 10 gram and travelling at a speed of 2 m/s.

Solution:

It is given that mass of object before explosion is,m = 10 g

Speed of object before explosion, v = 2 m/s

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Let $v_1, v_2 \text{ and}\ v_3$ be the velocities of the three fragments.

Therefore, according to the law of conservation of momentum,

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So the x- component of the velocity of the m2 fragment after the explosion is,

$3v_{2x} = 20$

∴ $v_{2x} = 6.67 \ m/s$

6 0
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