Ask question

Ask question

All categories

2 years ago

13

Mo is on a baseball team and hears that a ball thrown at a 45 degree angle from the ground will travel the furthest distance. Ho

w should Mo release the ball for the furthest travel?

1 answer:

Galina-37 [17]2 years ago

3 0

Answer:

Explanation:

Usually the angle between the y axis and x axis is 90° and we know that for furthest travel the degree angle must be 45° with the horizontal, Mo must release the ball about halfway between straight ahead and straight up

You might be interested in

A physics student stands on the rim of the canyon and drops a rock. The student measures the time for it to reach the bottom to

madam [21]

Answer:

Canyon is 50.176 meter deep.

Explanation:

The students is standing on the rim of the canyon and drops down a rock from the rim(cliff). We have to find what is the depth of the canyon i.e. how much below is the ground from the cliff.

Given data:

Time = t = 3.2 s

Initial velocity = $v_{i}$ = 0 m/s

Gravitational acceleration = g = 9.8 m/s²

Height = h = ?

According to second equation of motion

$h = v_{i}t + \frac{1}{2}gt^{2}$

As initial velocity is zero, So the first term of right hand side of above equation equal to zero

$h = \frac{1}{2}gt^{2}$

h = (0.5)(9.8)(3.2)²

h = 50.176 m

This means, the rock traveled 50.176 meters to reach the bottom of the Canyon. So, the Canyon is 50.176 meter deep.

5 0

2 years ago

In springboard diving, the diver strides out to the end of the board, takes a jump onto its end, and uses the resultant spring-l

Ksenya-84 [330]

Answer:

10.4 m/s

Explanation:

The problem can be solved by using the following SUVAT equation:

$v=u+at$

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

For the diver in the problem, we have:

$u=+6.3 m/s$ is the initial velocity (positive because it is upward)

$a=g=-9.8 m/s^2$ is the acceleration of gravity (negative because it is downward)

By substituting t = 1.7 s, we find the velocity when the diver reaches the water:

$v=+6.3 + (-9.8)(1.7)=-10.4 m/s$

And the negative sign means that the direction is downward: so, the speed is 10.4 m/s.

3 0

2 years ago

Read 2 more answers

An ant is crawling along a yardstick that is pointed with the 0-inch mark to the east and the 36-inch mark to the west. It start

FrozenT [24]

Answer:

The total distance traveled is 28 inches.
The displacement is 2 inches to the east.

Explanation:

Lets put a frame of reference in the problem. Starting the frame of reference at the point with the 0-inch mark, and making the unit vector $\hat{i}$ pointing in the west direction, the ant start at position

$\vec{r}_0 = 16 \ inch \ \hat{i}$

Then, moves to

$\vec{r}_1 = 29 \ inch \ \hat{i}$

so, the distance traveled here is

$d_1 = |\vec{r}_1 - \vec{r}_0 | = | 29 \ inch \ \hat{i} - 16 \ inch \ \hat{i} |$

$d_1 = | 13 \ inch \ \hat{i} |$

$d_1 = 13 \ inch$

after this, the ant travels to

$\vec{r}_2 = 14 \ inch \ \hat{i}$

so, the distance traveled here is

$d_2 = |\vec{r}_2 - \vec{r}_1 | = | 14 \ inch \ \hat{i} - 29 \ inch \ \hat{i} |$

$d_2 = | - 15 \ inch \ \hat{i} |$

$d_2 = 15 \ inch$

The total distance traveled will be:

$d_1 + d_2 = 13 \ inch + 15 \ inch = 28 \ inch$

The displacement is the final position vector minus the initial position vector:

$\vec{D}=\vec{r}_2 - \vec{r}_1$

$\vec{D}= 14 \ inch \ \hat{i} - 16 \ inch \ \hat{i}$

$\vec{D}= - 2 \ inch \ \hat{i}$

This is 2 inches to the east.

6 0

2 years ago

The air surrounding an airplane in flight exerts a drag force that acts opposite to the airplane's motion. When an Airbus A380 i

Rainbow [258]

Answer:

$4.32\cdot 10^5 hp$ , $3.22\cdot 10^8 W$

Explanation:

The jet is flying at constant velocity: this means that its acceleration is zero, so the net force acting on the jet is also zero.

Therefore, we can write:

$4F_T - F_d = 0$

where

$F_T = 322,000 N$ is the thrust force generated by each engine of the jet

$F_d$ is the drag force

Solving for Fd,

$F_d = 4 F_T = 4(322,000)=1.288\cdot 10^6 N$

The velocity of the jet is

$v=250 m/s$

So, the rate at which the drag force does work (which is the power) is

$P=F_d v$

and substituting

$F_d = 1.288\cdot 10^6 N\\v = 250 m/s$

we find

$P=(1.288\cdot 10^6)(250)=3.22\cdot 10^8 W$

Converting into horsepower,

$P=\frac{3.22\cdot 10^8}{746}=4.32\cdot 10^5 hp$

4 0

2 years ago

The flat-bed trailer carries two 1500-kg beams with the upper beam secured by a cable. The coefficients of static friction betwe

Novosadov [1.4K]

Answer:

a) a= 8.33 m/s², T = 12.495 N , b) a = 2.45 m / s²

Explanation:

a) this is an exercise of Newton's second law. As the upper load is secured by a cable, it cannot be moved, so the lower load is determined by the maximum acceleration.

We apply Newton's second law to the lower charge

fr₁ + fr₂ = ma

The equation for the force of friction is

fr = μ N

Y Axis

N - W₁ –W₂ = 0

N = W₁ + W₂

N = (m₁ + m₂) g

Since the beams are the same, it has the same mass

N = 2 m g

We replace

μ₁ 2mg + μ₂ mg = m a

a = (2μ₁ + μ₂) g

a = (2 0.30 + 0.25) 9.8

a= 8.33 m/s²

Let's look for cable tension with beam 2

T = m₂ a

T = 1500 8.33

T = 12.495 N

b) For maximum deceleration the cable loses tension (T = 0 N), so as this beam has less friction is the one that will move first, we are assuming that the rope is horizontal

fr = m₂ a₂

N- w₂ = 0

N = W₂ = mg

μ₂ mg = m a₂

a = μ₂ g

a = 0.25 9.8

a = 2.45 m / s²

4 0

2 years ago