Ask question

Ask question

All categories

kolbaska11 [484]

3 years ago

9

a golfer hits a golf ball with a velocity of 36.0 meters/second at an angle of 28.0. if the hang time of the golf ball is 33.4 s

econds, what is the range of the golf ball

1 answer:

Kobotan [32]3 years ago

6 0

Solving this using the time, we know that range = horizontal velocity x time of flight

since there are no horizontal forces acting on the ball, there are no horizontal accelerations and the initial horizontal velocity of 36 cos 28 will be constant throughout. If we use the correct time of flight given the launch parameters, we have

range = 36 cos 28 x 3.44 s = 109.3 m

You might be interested in

An automobile traveling 95 km/h overtakes a 1.30-km-long train traveling in the same direction on a track parallel to the road.

SCORPION-xisa [38]

Answer: 6.175 km

Explanation:

from the question, we have the following

velocity of the automobile = 95 km/g

velocity of the train = 75 km/h

length of the train = 1.30 km

since the automobile and the train are moving in the same direction, we need to find the velocity of the car relative to the train which will be their difference in speed = 95 - 75 = 20 km/h

we need to find the time it takes the automobile to overtake the train using the formula time = distance / speed , with the distance being the length of the train.

time (t) = 1.3 / 20

= 0.065 hour

now we can find the distance traveled by the automobile using the the time taken for it to overtake the train and the speed of the automobile.

therefore, distance = speed x time

distance = 95 x 0.065 =6.175 km

5 0

4 years ago

a man drags a 8.10 kg bag of mulch at a constant speed, applying a 29.5 N at 38°. what is the coefficient of friction?

lesantik [10]

Answer:

The coefficient of friction is 0.38.

Explanation:

The free body diagram is drawn below.

Let $f$ be frictional force acting in the backward direction as shown. Let the coefficient of friction be $\mu$ . Let $N$ be the normal reaction force acting on the bag.

Given:

Mass of the bag is, $m=8.10\textrm{ kg}$

Force acting at $\theta = 38$ ° is $F= 29.5\textrm{ N}$

Acceleration due to gravity is, $g=9.8\textrm{ }m/s^{2}$

The force F can be resolved into its components as $F_{x}=F \cos \theta$ and $F_{y}=F \sin \theta$

Therefore,

$F_{x}=29.5\cos(38)=23.25\textrm{ N}\\F_{y}=29.5\sin(38)=18.16\textrm{ N}$

Now, as there is no acceleration in vertical direction, therefore,

Sum of upward forces = Sum of downward forces

$N+F_{y}=mg\\N=mg-F_{y}=8.10\times 9.8-18.16\\N=79.38-18.16=61.22\textrm{ N}$

Now, as the bag is moving at a constant speed, so acceleration in the horizontal direction is also zero as acceleration is the rate of change of velocity.

Therefore, backward force = forward force.

$f=F_{x}\\f=23.25\textrm{ N}$

Now, frictional force is given as:

$f=\mu N\\\mu = \frac{f}{N}=\frac{23.25}{61.22}=0.38$

Therefore, the coefficient of friction is 0.38.

8 0

4 years ago

The mean distance of an asteroid from the Sun is 2.98 times that of Earth from the Sun. From Kepler's law of periods, calculate

lutik1710 [3]

Answer:

The asteroid requires 5.14 years to make one revolution around the Sun.

Explanation:

Kepler's third law establishes that the square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit:

$T^{2} = a^{3}$ (1)

Where T is the period of revolution and a is the semi-major axis.

In the other hand, the distance between the Earth and the Sun has a value of $1.50x10^{8} Km$ . That value can be known as well as an astronomical unit (1AU).

But 1 year is equivalent to 1 AU according with Kepler's third law, since 1 year is the orbital period of the Earth.

For the special case of the asteroid the distance will be:

$a = 2.98(1.50x10^{8}Km)$

$a = 4.47x10^{8}Km$

That distance will be expressed in terms of astronomical units:

$4.47x10^{8}Km.\frac{1AU}{1.50x10^{8}Km}$ ⇒ $2.98AU$

Finally, from equation 1 the period T can be isolated:

$T = \sqrt{a^{3}}$

$T = \sqrt{(2.98)^{3}}$

$T = \sqrt{26.463592}$

$T = 5.14AU$

Then, the period can be expressed in years:

$5.14AU.\frac{1yr}{1AU} ⇒ 5.14 yr$

$T = 5.14 yr$

Hence, the asteroid requires 5.14 years to make one revolution around the Sun.

8 0

3 years ago

Which of the following is an inelastic collision?

Luda [366]

I'm pretty sure B would be your answer.

7 0

3 years ago

Computers have become a familiar ___ of communication, connecting people around the world. a. mediation

vredina [299]

The correct answer is c. Medium!
Medium is a way of doing something, so computers a a way of communicating!
Other options:

a. Mediation is the process of reconciling two opposing sides
b. proponent is a person who is advocating for something.
d. exposition is for a example an explanation of a theory

7 0

3 years ago