harold threw a baseball 3 meters. harold is 2 meters tall. how fast did he throw the ball and how long was it in the air

You throw a glob of putty straight up toward the ceiling, which is 3.60 m above the point where the putty leaves your hand. The

Sliva [168]

Answer:

Explanation:

Given

Height of ceiling is $h=3.6\ m$

Initial speed of Putty $u=9.5\ m/s$

Speed of Putty just before it strike the ceiling is given by

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

$v^2-9.5^2=2\times (-9.8)\times 3.6$

$v^2=19.69$

$v=4.43\ m/s$

time taken by putty to reach the ceiling

$v=u+at$

$4.43=9.5-9.8\times t$

$t=\frac{5.07}{9.8}$

$t=0.517\ s$

8 0

3 years ago

A satellite circles the Earth in an orbit whose radius is twice the Earth’s radius. The Earth’s mass is 5.98 x 1024 kg, and its

gavmur [86]

Hello!

Recall the period of an orbit is how long it takes the satellite to make a complete orbit around the earth. Essentially, this is the same as 'time' in the distance = speed * time equation. For an orbit, we can define these quantities:

$d = 2\pi r$ ← The circumference of the orbit

speed = orbital speed, we will solve for this later

time = period

Therefore:

$T = \frac{2\pi r}{v}$

Where 'r' is the orbital radius of the satellite.

First, let's solve for 'v' assuming a uniform orbit using the equation:
$v = \sqrt{\frac{Gm}{r}}$

G = Gravitational Constant (6.67 × 10⁻¹¹ Nm²/kg²)

m = mass of the earth (5.98 × 10²⁴ kg)

r = radius of orbit (1.276 × 10⁷ m)

Plug in the givens:
$v = \sqrt{\frac{(6.67*10^{-11})(5.98*10^{24})}{(1.276*10^7)}} = 5590.983 m/s$

Now, we can solve for the period:

$T = \frac{2\pi (1.276*10^7)}{5590.983} =\boxed{ 14339.776 s}$

7 0

2 years ago

Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a larg

IRINA_888 [86]

Answer:

Time : 7.96 s

Distance Traveled : 357.8 m

Explanation:

In order to solve this problem, we first consider the accelerated motion of rocket. We will be using the subscript 1 for accelerated motion.

So, for accelerated motion, we have:

Acceleration = a₁ = 14.5 m/s²

Time Period = t₁ = 3.1 s

Initial Velocity = Vi₁ = 0 m/s (Since, it starts from rest)

Final Velocity = Vf₁

Distance covered by sled during acceleration motion = s₁

Now, using 1st equation of motion:

Vf₁ = Vi₁ + (a₁)(t₁)

Vf₁ = 0 m/s + (14.5 m/s²)(3.1 s)

Vf₁ = 44.95 m/s

Now, using 2nd equation of motion:

s₁ = (Vi₁)(t) + (0.5)(a₁)(t₁)

s₁ = (0 m/s)(3.1 s) + (0.5)(14.5 m/s²)(3.1 s)

s₁ = 22.5 m

Now, we first consider the decelerated motion of rocket. We will be using the subscript 2 for decelerated motion.

So, for accelerated motion, we have:

Deceleration = a₂ = - 5.65 m/s²

Time Period = t₂ = ?

Initial Velocity = Vi₂ = Vf₁ = 44.95 m/s (Since, decelerate motion starts, where accelerated motion ends)

Final Velocity = Vf₂ = 0 m/s (Since, rocket will eventually stop)

Distance covered by sled during deceleration motion = s₂

Now, using 1st equation of motion:

Vf₂ = Vi₂ + (a₂)(t₂)

0 m/s = 44.95 m/s + (- 5.65 m/s²)(t₂)

t₂ = (44.95 m/s)/(5.65 m/s²)

t₂ = 7.96 s

Now, using 2nd equation of motion:

s₂ = (Vi₂)(t₂) + (0.5)(a₂)(t₂)

s₂ = (44.95 m/s)(7.96 s) + (0.5)(- 5.65 m/s²)(7.96 s)

s₂ = 357.8 m - 22.5 m

s₂ = 335.3 m

Thus, the total distance covered by sled will be:

Total Dustance = S = s₁ + s₂

S = 22.5 m + 335.3 m

S = 357.8 m

7 0

3 years ago

A box, compressing a spring by 4 m, is loaded and ready. Moments later the 8.0 kg box has been launched with

Mrac [35]

Answer:

50 N/m

Explanation:

Elastic energy = kinetic energy

EE = KE

½ kx² = ½ mv²

½ k (4 m)² = ½ (8.0 kg) (10.0 m/s)²

k = 50 N/m

3 0

3 years ago

The table below shows different items that are found sitting on the counter in a kitchen. The temperature of the kitchen is 72 d

melisa1 [442]

This question is incomplete

Complete Question

The table below shows different items that are found sitting on the counter in a kitchen. The temperature of the kitchen is 72 degrees Fahrenheit. Classify the objects according to whether their molecules will speed up or slow down after being left in the kitchen for a period of time.

Drag the objects to the correct category.

Object Temperature (Fahrenheit)

ice cubes 26°F

glass of tea 60°F

cooked piece of meat 160°F

butter 55°F

pot of water 75°F

bowl of soup 140°F

Answer:

Molecules that would speed up

ice cubes 26°F

glass of tea 60°F

butter 55°F

Molecules that would slow down

cooked piece of meat 160°F

pot of water 75°F

bowl of soup 140°F

Explanation:

Molecules that are found or contained in a substances have the tendency to react in such as what that they are are sped up or slowed down. This is due to their exposure to various kinds of changes in temperature, which could be a hot temperature , cold temperature e.t.c

When a substance is exposed to a hot temperature, the molecules of that substance speed up depending on how hot the temperature is while when a substance is cooled down or exposed to cold temperature, the molecules tend to slow down.

In the question above, the temperature of the kitchen is 72 degrees Fahrenheit.

Molecules that would speed up after been left in the Kitchen for a while:

a) Ice cubes 26°F: This is because 72°F is a warm took temperature, so the ice cubes would melt causing the molecules to speed up.

b)Glass of tea 60°F: This glass of tea is at a cool temperature of 60°F, when it is kept in the Kitchen which has a temperature of 72°F for while, the temperature of the glass of tea would increase due to an increase in its rate of reaction causing the molecules of the glass of water to speed up.

c) Butter 55°F: This is because 72°F is a warm temperature, so the butter would melt, increasing its rate of reaction and causing the molecules to speed up

Molecules that would slow down

a) Cooked piece of meat 160°F : 160°F is a very hot temperature and when it is left in a the Kitchen(72°F) , the cooked meat would begin to cool down and drop in temperature from 160°F, causing the molecules to slow down.

b) Pot of water 75°F: The temperature of the kitchen and the temperature of the pot of water is the same, hence the molecules of the water would slow down.

c) Bowl of soup 140°F : This bowl of soup is very hot and when it is left in the Kitchen(72°F) , the bowl of soup would begin to cool down and drop in temperature from 140°F, slowing down the rate of reaction and causing the molecules to slow down.

3 0

3 years ago