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2 years ago

A object was fired at 17ms at 40 degree angle horizontal , if object traveled 29meters how long was it in the air

1 answer:

8 0

First you need to find the initial x component which is 13.02. After that you plug it into the 3rd formula on the AP physics formula sheet. Then rearrange it into the quadratic formula and solve for time. The answer should be 4.1 seconds.

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Can you help me please?

egoroff_w [7]

<h3>Answer</h3>

option B)

19N

<h3>Explanation</h3>

If the object is at equilibrium, then the net force acting upon the object should be 0 N. Thus, if all the forces are added together, horizontal and vertical forces separately, then the resultant force (the vector sum) should be 0 Newton.

As we only need to find the magnitude of x-component of force F

so find all x component/horizontal forces acting on the object.

50cos(40) - 40cos(25) + 30cos(55) + x = 0

38.30 - 36.25 + 17.21 + x + = 0

19.26 + x = 0

x = - 19.26

x ≈ 19 (magnitude only)

7 0

4 years ago

You're carrying a 3.0-m-long, 24 kg pole to a construction site when you decide to stop for a rest. You place one end of the pol

kozerog [31]

Answer:

$F=133N$

Explanation:

From the question we are told that:

Length $l=3.0m$

Mass $m=24kg$

Distance from Tip $d=35cm$

Generally, the equation for Torque Balance is mathematically given by

$mg(l/2)=F(l-d)$

$2*9.81(3/2)=F(3-35*10^-2)$

Therefore

$F=133N$

8 0

3 years ago

A +27 nCnC point charge is placed at the origin, and a +6 nCnC charge is placed on the xx axis at x=1mx=1m. At what position on

svet-max [94.6K]

Answer:

The position on the x axis is 0.32 m.

Explanation:

Given that,

Point charge = 27 nC

Charge = 6 nC

Distance = 1

We need to calculate the distance

Using formula of electric field

$\dfrac{kq_{1}}{x^2}=\dfrac{kq_{2}}{(r-x)^2}$

Put the value into the formula

$\dfrac{27\times10^{-9}}{x^2}=\dfrac{6\times10^{-9}}{(1-x)^2}$

$\dfrac{27}{x^2}=\dfrac{6}{(1-x)^2}$

$\dfrac{(1-x)^2}{x^2}=\dfrac{27}{6}$

$\dfrac{1-x}{x}=\sqrt{\dfrac{27}{6}}$

$\dfrac{1}{x}=\sqrt{\dfrac{27}{6}}+1$

$x=0.32\ m$

Hence, The position on the x axis is 0.32 m.

5 0

3 years ago

A stubborn, 100 kgkg mule sits down and refuses to move. To drag the mule to the barn, the exasperated farmer ties a rope around

jolli1 [7]

Answer:

No, the farmer is not able to move the mule.

Explanation:

Mass =100 kg

Force=F=800 N

The coefficient between the mule and the ground= $\mu_s=0.8$

$\mu_k=0.5$

Static friction force,f= $\mu N$

Normal force=N=mg

Static friction force,f= $\mu_s mg=0.8\times 100\times 9.8=784 N$

Using $g=9.8m/s^2$

F<f

Static friction force is greater than applied force.

Therefore , the farmer is not able to move the mule.

4 0

3 years ago

A projectile is launched horizontally from a height of 8.0 m. The projectile travels 6.5 m before hitting the ground.

Effectus [21]

Time it takes the projectile to hit the ground after being thrown up:

√h/1/2a

√8/(.5)(9.81)

√8/4.905

√1.630988787

= 1.277101714

= 1. 28

hope this helps :)

7 0

3 years ago