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3 years ago

Resistance question for physics!!!

Physics

1 answer:

7nadin3 [17]3 years ago

4 0

Answer:

Explanation:

all the resistance numbers added together 2.0+6.0+10.0+4.0

resistance formula : v bettary = v1 + v2 + v3 ....

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A basketball is shot at 14.0 m/s at a 65.0 degree angle. What is the magnitude only (no direction) of the velocity of the ball 2

Alex787 [66]

Answer:

10.4 m/s

Explanation:

I just know it's right!

8 0

3 years ago

A cannon fires a cannonball forward at a velocity of 48.1 m/s horizontally. If the cannon is on a mounted wagon 1.5 m tall, how

GalinKa [24]

Answer: 473.640 m

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the cannonball has two components: x-component and y-component. Being their main equations as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$V_{o}=48.1 m/s$ is the cannonball's initial velocity

$\theta=0$ because we are told the cannonball is shot horizontally

$t$ is the time since the cannonball is shot until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=1.5m$ is the initial height of the cannonball

$y=0$ is the final height of the cannonball (when it finally hits the ground)

$g=9.8m/s^{2}$ is the acceleration due gravity

We need to find how far (horizontally) the cannonball has traveled before landing. This means we need to find the maximum distance in the x-component, let's call it $X_{max}$ and this occurs when $y=0$ .

So, firstly we will find the time with (2):

$0=1.5 m+48. 1 m/s sin(0\°) t-(4.9 m/s^{2})t^2$ (3)

Rearranging the equation:

$0=-(4.9 m/s^{2})t^2+48. 1 m/s sin(0\°) t+1.5 m$ (4)

$-(4.9 m/s^{2})t^2+(48. 1 m/s) t+1.5 m=0$ (5)

This is a quadratic equation (also called equation of the second degree) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ (6)

Where:

$a=-4.9 m/s^{2}$

$b=48.1 m/s$

$c=1.5 m$

Substituting the known values:

$t=\frac{-48.1 \pm \sqrt{48.1^{2}-4(-4.9)(1.5)}}{2(-4.9)}$ (7)

Solving (7) we find the positive result is:

$t=9.847 s$ (8)

Substituting this value in (1):

$x=(48.1 m/s)cos(0\°) (9.847 s)$ (9)

$x=473.640 m$ This is the horizontal distance the cannonball traveled before it landed on the ground.

3 0

3 years ago

An athlete does one push-up. In the process, she moves half of her body weight, 250 newtons, a distance of 20 centimeters. This

s2008m [1.1K]

We know that an athlete moves half of her body weight 250 N a distance of 20 cm.
So: F = 250 N, d = 20 cm = 0.2 m
The work formula:
W = F * d
W = 250 N * 0.2 m
W = 50 J
Answer: Her work after one push-up is 50 J.

6 0

3 years ago

A moving walkway has a speed of 0.9 m/s to the east. A stationary observer sees a man walking on the walkway with a velocity of

Leona [35]

B or .2 East is the correct answer

3 0

3 years ago

Read 2 more answers

How does a parallel circuit change when a branch is added?

Leto [7]

Answer: The total resistance increases, so the current in the circuit decreases.

3 0

2 years ago