A blank is a statement the summerizes a pattern found in nature

A ski jumper starts with a horizontal take-off velocity of 25 m/s and lands on a straight landing hill inclined at 30o. Determin

melamori03 [73]

Answer:

0.34s, 8.5m,31.89m

Explanation:

The above motion defines a projectile motion.

Now the athletes lands on a cliff 30° to the horizontal this means the velocity at that point would be 25m/s cos30°

Now from Newton's law of motion.

The body would be decelerating so,

V = u - gt

Where u is initial velocity and v is final velocity. g is acceleration of free fall due to gravity.

Hence,

V-U/ -g = t

Hence 25cos30 - 25/ -9.8 = 0.34s.

2.Now the length of the jump is defined as the total horizontal distance which is marked off by the horizontal velocity and time taken for take off and landing.

Hence Distance,S = u × t

25 ×0.34 =8.5m.

3. The maximum height is defined that at that point the Final velocity is 0m/s

Now the initial velocity is 25m/s

From Newton's law that;

V2= U2 -2gH; where U and V are initial and final velocity and H is height.

Hence H = V2-U2/-2g

=(0)^2- (25)^2/ -2×9.8

= -625/-19.6 =31.89m

8 0

3 years ago

A rock is suspended by a light string. When the rock is in air, the tension in the string is 51.9 N . When the rock is totally i

Luden [163]

Answer:

$\rho _{liquid}=1995.07kg/m^{3}$

Explanation:

When the rock is immersed in unknown liquid the forces that act on it are shown as under

1) Tension T by the string

2) Weight W of the rock

3) Force of buoyancy due to displaced liquid B

For equilibrium we have $T_{3}+B = W_{rock}$

$T_{3}+\rho _{Liquid}V_{rock}g$ = $W_{rock}.....(\alpha)$

When the rock is suspended in air for equilibrium we have

$T_{1}=W_{rock}....(\beta)$

When the rock is suspended in water for equilibrium we have

$T_{2}$ + $\rho _{water}V_{rock}g$ = $W_{rock}.....(\gamma)$

Using the given values of tension and solving α,β,γ simultaneously for $\rho _{Liquid}$ we get

$W_{rock}=51.9N\\31.6+1000\times V_{rock}\times g=51.9N\\\\11.4+\rho _{liquid}V_{rock}g=51.9N\\\\$

Solving for density of liquid we get

$\rho _{liquid}=\frac{51.9-11.4}{51.9-31.6}\times 1000$

$\rho _{liquid}=1995.07kg/m^{3}$

5 0

3 years ago

A rocket's acceleration is 6.0 m/s 2 . Assuming it starts at 0 m/s, how long will it take for the rocket to reach a velocity of

kupik [55]

At=v-u, where u=0

t= v/a=42/6 = 7
Answer: 7 seconds

8 0

3 years ago

Ten seconds after starting from rest, a freely-falling object will have a speed of about ________________________

yaroslaw [1]

Answer:

100 m/s

Explanation:

Speed increases at a rate of 10 m/s (actually 9.8 m/s) every second. Thus after 10 seconds, the speed is 10 x 10 = 100 m/s.

6 0

3 years ago

An object is launched at a velocity of 20 m/s in a direction making an angle of 25° upward with the horizontal.

Hitman42 [59]

Answer:

(a) max. height = 3.641 m

(b) flight time = 1.723 s

(c) horizontal range = 31.235 m

(d) impact velocity = 20 m/s

Above values have been given to third decimal. Adjust significant figures to suit accuracy required.

Explanation:

This problem requires the use of kinematics equations

v1^2-v0^2=2aS .............(1)

v1.t + at^2/2 = S ............(2)

where

v0=initial velocity

v1=final velocity

a=acceleration

S=distance travelled

SI units and degrees will be used throughout

Let

theta = angle of elevation = 25 degrees above horizontal

v=initial velocity at 25 degrees elevation in m/s

a = g = -9.81 = acceleration due to gravity (downwards)

(a) Maximum height

Consider vertical direction,

v0 = v sin(theta) = 8.452 m/s

To find maximum height, we find the distance travelled when vertical velocity = 0, i.e. v1=0,

solve for S in equation (1)

v1^2 - v0^2 = 2aS

S = (v1^2-v0^2)/2g = (0-8.452^2)/(2*(-9.81)) = 3.641 m/s

(b) total flight time

We solve for the time t when the vertical height of the object is AGAIN = 0.

Using equation (2) for vertical direction,

v0*t + at^2/2 = S substitute values

8.452*t + (-9.81)t^2 = 3.641

Solve for t in the above quadratic equation to get t=0, or t=1.723 s.

So time for the flight = 1.723 s

(c) Horiontal range

We know the horizontal velocity is constant (neglect air resistance) at

vh = v*cos(theta) = 25*cos(25) = 18.126 m/s

Time of flight = 1.723 s

Horizontal range = 18.126 m/s * 1.723 s = 31.235 m

(d) Magnitude of object on hitting ground, Vfinal

By symmetry of the trajectory, Vfinal = v = 20, or

Vfinal = sqrt(v0^2+vh^2) = sqrt(8.452^2+18.126^2) = 20 m/s

7 0

4 years ago