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3 years ago

Using the projectiles equations for y and x at what second does the ball have a vertical velocity of -39.2 m/s^2

Physics

1 answer:

vovangra [49]3 years ago

8 0

Answer:

Explanation:

In projectile the time taken by an object to reach its maximum height is the maximum time (tmax).

tmax = Usin $\theta$ /g

U = velocity of the ball

$\theta$ = angle of projection = 90° (vertical velocity)

g = acceleration due to gravity = 9.8m/s²

tmax = Usin90°/9.8

tmax = U/9.8

T the maximum height U = Uy = 39.2 m/s²

tmax = 39.2/9.8

tmax = 4secs

To calculate the time at which the ball will have an horizontal velocity, we will use the same formula but $\theta$ will be 0° in this case

If Ux = Ucos $\theta$

8 = Ucos0

U = 8m/s

Since t = Usin $\theta$ /g

t = 8sin0/9.8

t = 0sec

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A bar of length L = 8 ft and midpoint D is falling so that, when θ = 27°, ∣∣v→D∣∣=18.5 ft/s , and the vertical acceleration of p

777dan777 [17]

Answer:

alpha=53.56rad/s

a=5784rad/s^2

Explanation:

First of all, we have to compute the time in which point D has a velocity of v=23ft/s (v0=0ft/s)

$v=v_0+at\\\\t=\frac{v}{a}=\frac{(23\frac{ft}{s})}{32.17\frac{ft}{s^2}}=0.71s$

Now, we can calculate the angular acceleration (w0=0rad/s)

$\theta=\omega_0t +\frac{1}{2}\alpha t^2\\\alpha=\frac{2\theta}{t^2}$

$\alpha=\frac{27}{(0.71s)^2}=53.56\frac{rad}{s^2}$

with this value we can compute the angular velocity

$\omega=\omega_0+\alpha t\\\omega = (53.56\frac{rad}{s^2})(0.71s)=38.02\frac{rad}{s}$

and the tangential velocity of point B, and then the acceleration of point B:

$v_t=\omega r=(38.02\frac{rad}{s})(4)=152.11\frac{ft}{s}\\a_t=\frac{v_t^2}{r}=\frac{(152.11\frac{ft}{s})^2}{4ft}=5784\frac{rad}{s^2}$

hope this helps!!

6 0

3 years ago

Read 2 more answers

A spy satellite uses a telescope with a 1.7-m-diameter mirror. It orbits the earth at a height of 180 km.

WINSTONCH [101]

Answer: the minimum spacing that must be there between two objects on the earth's surface if they are to be resolved as distinct objects by this telescope 6.45 cm

Explanation:

Given that;

diameter of the mirror d = 1.7 m

height h = 180 km = 180 × 10³ m

wavelength λ = 500 nm = 5 × 10⁻⁹ m

Now Angular separation from the peak of the central maximum is expressed as;

sin∅= 1.22 λ / d

sin∅ = (1.22 × 5 × 10⁻⁹) / 1.7

sin∅ = 3.588 × 10⁻⁷

we know that;

sin∅ = object separation / distance from telescope

object separation = sin∅ × distance from telescope

object separation = 3.588 × 10⁻⁷ × 180 × 10³

object separation =6.45 × 10⁻² m

then we convert to centimeter

object separation = 6.45 cm

Therefore the minimum spacing that must be there between two objects on the earth's surface if they are to be resolved as distinct objects by this telescope 6.45 cm

5 0

3 years ago

Calculate the force needed to accelerate a car of a mass 1000 kg by 3 m/s2

beks73 [17]

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 1000 × 3

We have the final answer as

Hope this helps you

5 0

3 years ago

Why would a flat sheet of paper and a wad of paper with the same mass not fall through the air at the same rate?

Gelneren [198K]

The flat sheet of paper has more surface area than the crumpled ball

7 0

3 years ago

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At a circus, a human cannonball is shot from a cannon at 15m/s at an angle of 40° above horizontal. She

Citrus2011 [14]

Answer:

The height is $H = 6.74 \ m$

The horizontal distance is $D = 23.74 \ m$

Explanation:

From the question we are told that

The speed is $v = 15 \ m/s$

The angle is $\theta = 40^o$

The height of the cannon from the ground is h = 2 m

The distance of the net from the ground is k = 1 m

Generally the maximum height she reaches is mathematically represented as

$H = \frac{v^2 sin^2 \theta }{2 * g } + h$

=> $H = \frac{(15)^2 [sin (40)]^2 }{2 * 9.8} + 2$

=> $H = 6.74 \ m$

Generally from kinematic equation

$s = ut + \frac{1}{2} at^2$

Here s is the displacement which is mathematically represented as

s = [-(h-k)]

=> s = -(2-1)

=> s = -1 m

There reason why s = -1 m is because upward motion canceled the downward motion remaining only the distance of the net from the ground which was covered during the first half but not covered during the second half

a = -g = -9.8

$u = v sin (\theta)$