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jarptica [38.1K]

3 years ago

10

Two vectors, X and Y, form a right angle. Vector X is 48 inches long and vector Y is 14 inches long. the length of the resultant

vector is____ inches.

1 answer:

Scorpion4ik [409]3 years ago

4 0

Solution is provided below hope this helps

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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

3 years ago

To test the quality of a tennis ball, you drop it onto the floor from a height of 4.00 m. It rebounds to a height of 2.00 m. If

arlik [135]

Answer:

Part a)

$a = 1260.3 m/s^2$

Part b)

Direction = upwards

Explanation:

When ball is dropped from height h = 4.0 m

then the speed of the ball just before it will strike the ground is given as

$v_f^2 - v_i^2 = 2 a d$

$v_1^2 - 0^2 = 2(9.81)(4.0)$

$v_1 = 8.86 m/s$

Now ball will rebound to height h = 2.00 m

so the velocity of ball just after it will rebound is given as

$v_f^2 - v_i^2 = 2 a d$

$0 - v_2^2 = 2(-9.81)(2.00)$

$v_2 = 6.26 m/s$

Part a)

Average acceleration is given as

$a = \frac{v_f - v_i}{\Delta t}$

$a = \frac{6.26 - (-8.86)}{12.0 \times 10^{-3}}$

$a = 1260.35 m/s^2$

Part B)

As we know that ball rebounds upwards after collision while before collision it is moving downwards

So the direction of the acceleration is vertically upwards

7 0

3 years ago

A projectile is fired vertically upwards and reaches a height of 78.4 m. Find the velocity of projection and the time it takes t

Musya8 [376]

Answer:

1.) U = 39.2 m/s

2.) t = 4s

Explanation: Given that the

height H = 78.4m

The projectile is fired vertically upwards under the acceleration due to gravity g = 9.8 m/s^2

Let's assume that the maximum height = 78.4m. And at maximum height, final velocity V = 0

Velocity of projections can be achieved by using the formula

V^2 = U^2 - 2gH

g will be negative as the object is moving against the gravity

0 = U^2 - 2 × 9.8 × 78.4

U^2 = 1536.64

U = sqrt( 1536.64 )

U = 39.2 m/s

The time it takes to reach its highest point can be calculated by using the formula;

V = U - gt

Where V = 0

Substitute U and t into the formula

0 = 39.2 - 9.8 × t

9.8t = 39.2

t = 39.2/9.8

t = 4 seconds.

7 0

4 years ago

A 100-kg running back runs at 5 m/s into a stationary linebacker. It takes 0.5 s for the running back to be completely stopped.

Elza [17]

Answer:

1000 N

Explanation:

First, we need to find the deceleration of the running back, which is given by:

$a=\frac{v-u}{t}$

where

v = 0 is his final velocity

u = 5 m/s is his initial velocity

t = 0.5 s is the time taken

Substituting, we have

$a=\frac{0-5 m/s}{0.5 s}=-10 m/s^2$

And now we can calculate the force exerted on the running back, by using Newton's second law:

$F=ma=(100 kg)(-10 m/s^2)=-1000 N$

so, the magnitude of the force is 1000 N.

6 0

3 years ago

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How and why will my weight change if i took a trip to the moon

MatroZZZ [7]

weight less on moon than on earth.

high on lift off - G force

low in orbit.

zero at a point between earth and moon

7 0

3 years ago

Read 2 more answers