Question 3 of 5

You step onto a hot beach with your bare feet. A nerve impulse, generated in your foot, travels through your nervous system at a

barxatty [35]

Answer:16.3

Explanation:the

Explanation is in words

on a pic and i showed extra proof

3 0

2 years ago

What is the speed of the object at the end of 10 s?

Ostrovityanka [42]

Answer:

-100m/s not shure tho thx tho

6 0

2 years ago

31. If you threw a baseball straight out at 45 m/s from a height of 1.5 meters (A) how long would it be in the air? B) How far o

coldgirl [10]

Answer:

A) t = 0.55 s

B) x = 24.8 m

Explanation:

A) We can find the time at which the ball will be in the air using the following equation:

$y_{f} = y_{0} + v_{0y}t - \frac{1}{2}gt^{2}$

Where:

$y_{f}$ is the final height= 0

$y_{0}$ is the initial height= 1.5 m

$v_{0y}$ is the component of the initial speed in the vertical direction = 0 m/s

t: is the time =?

g: is the gravity = 9.81 m/s²

$0 = 1.5 m - \frac{1}{2}9.81 m/s^{2}t^{2}$

By solving the above equation for t we have:

$t = \sqrt{\frac{2*1.5 m}{9.81 m/s^{2}}} = 0.55 s$

Hence, the ball will stay 0.55 seconds in the air.

B) We can find the distance traveled by the ball as follows:

$x_{f} = x_{0} + v_{0x}t + \frac{1}{2}at^{2}$

Where:

a: is the acceleration in the horizontal direction = 0

$x_{f}$ is the final position =?

$x_{0}$ is the initial position = 0

$v_{0x}$ is the component of the initial speed in the horizontal direction = 45 m/s

$x_{f} = x_{0} + v_{0x}t + \frac{1}{2}at^{2}$

$x_{f} = 0 + 45 m/s*0.55 s + 0 = 24.8 m$

Therefore, the ball will travel 24.8 meters.

I hope it helps you!

3 0

2 years ago

Please help on this one

Sphinxa [80]

Since this is a projectile motion problem, break down each of the five kinematic quantities into x and y components. To find the range, we need to identify the x component of the displacement of the ball.

Let's break them down into components.

X Y

v₁ 32 cos50 m/s 32 sin50 m/s

v₂ 32 cos50 m/s ?

Δd ? 0

Δt ? ?

a 0 -9.8 m/s²

Let's use the following equation of uniform motion for the Y components to solve for time, which we can then use for the X components to find the range.

Δdy = v₁yΔt + 0.5ay(Δt)²

0 = v₁yΔt + 0.5ay(Δt)²

0 = Δt(v₁ + 0.5ayΔt), Δt ≠ 0

0 = v₁ + 0.5ayΔt

0 = 32sin50m/s + 0.5(-9.8m/s²)Δt

0 = 24.513 m/s - 4.9m/s²Δt

-24.513m/s = -4.9m/s²Δt

-24.513m/s ÷ 4.9m/s² = Δt

5.00s = Δt

Now lets put our known values into the same kinematic equation, but this time for the x components to solve for range.

Δdₓ = v₁ₓΔt + 0.5(a)(Δt)²

Δdₓ = 32cos50m/s(5.00s) + 0.5(0)(5.00)²

Δdₓ = 32cos50m/s(5.00s)

Δdₓ = 102.846

Therefore, the answer is A, 102.9m. According to significant digit rules, neither would be correct, but 103m is the closest to 102.9m so I guess that is what it is.

6 0

3 years ago

A room with dimensions 7.00m×8.00m×2.50m is to be filled with pure oxygen at 22.0∘C and 1.00 atm. The molar mass of oxygen is 32

VashaNatasha [74]

1) 5765 mol

First of all, we need to find the volume of the gas, which corresponds to the volume of the room:

$V=7.00 m\cdot 8.00 m\cdot 2.50 m=140 m^3$