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icang [17]
3 years ago
15

A stone is thrown at an angle of 30 degrees above the horizontal from the top edge of a cliff with an initial speed of 12 m/s. A

stop watch measures the stone’s trajectory time from top of cliff to bottom to be 4.8s. What is the height of the cliff?
Physics
1 answer:
natta225 [31]3 years ago
5 0

v = initial velocity of launch of the stone = 12 m/s

θ = angle of the velocity from the horizontal = 30

Consider the motion of the stone along the vertical direction taking upward direction as positive and down direction as negative.

v₀ = initial velocity along vertical direction = v Sinθ = 12 Sin30 = 6 m/s

a = acceleration of the stone = - 9.8 m/s²

t = time of travel = 4.8 s

Y = vertical displacement of stone = vertical height of the cliff = ?

using the kinematics equation

Y = v₀ t + (0.5) a t²

inserting the values

Y = 6 (4.8) + (0.5) (- 9.8) (4.8)²

Y = - 84.1 m

hence the height of the cliff comes out to be 84.1 m

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Two cars collide inelastically and stick together after the collision. Before the collision, the magnitudes of their momenta are
dolphi86 [110]

Answer:

a) P1+P2

Explanation:

The magnitude of their combined momentum is just the addition of each momentum, because in this case of inelastic collision, the kinetic energy of the two cars are both converted to some form of energy because the velocity of both cars becomes zero, i.e., V=0, making P = mv = 0, this means the magnitude of P1 + P2 = 0.

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3 years ago
What is the minimum speed must a salmon jumping at an angle of 35.2 leave the water in m/s?
yuradex [85]

Height of the waterfall is 0.449 m

its horizontal distance will be 2.1 m

now let say his speed is v with which he jumped out so here the two components of his velocity will be

v_x = vcos35.2 = 0.817 v

v_y = vsin35.2 = 0.576 v

here the acceleration due to gravity is 9.81 m/s^2 downwards

now we can find the time to reach the other end by y direction displacement equation

\Delta y = v_y * t + \frac{1}{2} at^2

-0.449 = 0.576 * v *t - \frac{1}{2}*9.81 * t^2

also from x direction we can say

\Delta x = v_x * t

2.1 = 0.817 v* t

now we have

v* t = 2.57

we will plug in this value into first equation

- 0.449 = 0.576 * 2.57 - 4.905 * t^2

1.93 = 4.905 * t^2

t = 0.63 s

now as we know that

v* t = 2.57

t = 0.63 s

v = \frac{2.57}{0.63}

v = 4.1 m/s

so his minimum speed of jump is 4.1 m/s

8 0
3 years ago
5. The rate at which an organism uses energy, measured in humans at complete<br> rest.
yuradex [85]

Answer:

n. Abbr. BMR. The rate at which energy is used by an organism at complete rest, measured in humans by the heat given off per unit time, and expressed as the calories released per kilogram of body weight or per square meter of body surface per hour.

Explanation:

3 0
2 years ago
If you place 48 cm object 15 m in front of a convex mirror of focal length 5 m, what is the magnification of the image?
Arisa [49]

Answer:

m = 0.25

Explanation:

Given that,

Object distance, u = -15cm

Height of the object, h = 48

Focal length, f = cm

We need to find the magnification of the image.

Let v is the image distance. Using mirror's equation.

\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}\\\\=\dfrac{1}{5}-\dfrac{1}{(-15)}\\\\=3.75\ cm

Magnification,

m=\dfrac{-v}{u}\\\\=\dfrac{-3.75}{-15}\\\\=0.25

Hence, the magnification of the image is 0.25.

7 0
3 years ago
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