1. Calculate the height of tree, 250 m away that produces

FORCE AND DISPLACEMENT AT AN ANGLE A sailor pulls a boat a distance of 30.0 m along a dock using a rope that makes a 25.0° angle

Bingel [31]

Answer: 6117.58 J

Explanation:

We know that W=Fd*cos(theta) where theta is the angle between the displacement and the force.

In this case, we are given that F=225 N, d=30 m, and theta=25 degrees.

Plugging all this in we get

W=225*30*cos(25)=6117.58 J

7 0

3 years ago

A plane flies at 400 north of East for 150 miles,then flies 200 miles at an angle of 150 west of North. What is the plane's fina

ivanzaharov [21]

Answer:

341.46miles

Explanation:

Find the diagram attachment.

To get the displacement D, we will use the cosine rule as shown;

D² = 200²+150²-2(160)(400)cos65°

D² = 40000+22500-128000cos65°

D² = 62500+54095.14

D² = 116595.14

D = √116595.14

D= 341.46 miles

Hence the plane final displacement is 341.46miles

3 0

2 years ago

Define what is energy and work

Mashutka [201]

The work-energy principle states that an increase in the kinetic energy of a rigid body is caused by an equal amount of positive work done on the body by the resultant force acting on that body. Conversely, a decrease in kinetic energy is caused by an equal amount of negative work done by the resultant force.

8 0

3 years ago

Two point charges are fixed on the y axis: a negative point charge q1 = -25 μC at y1 = +0.18 m and a positive point charge q2 at

dedylja [7]

Answer:

$50.91 \mu C$

Explanation:

The magnitude of the net force exerted on q is known, we have the values and positions for $q_{1}$ and q. So, making use of coulomb's law, we can calculate the magnitude of the force exerted by $q_{1}$ on q. Then we can know the magnitude of the force exerted by $q_{2}$ about q, finally this will allow us to know the magnitude of $q_{2}$

$q_{1}$ exerts a force on q in +y direction, and $q_{2}$ exerts a force on q in -y direction.

$F_{1}=\frac{kq_{1} q }{d^2}\\F_{1}=\frac{(8.99*10^9)(25*10^{-6}C)(8.4*10^{-6}C)}{(0.18m)^2}=58.26 N\\$

The net force on q is:

$F_{T}=F_{1} - F_{2}\\25N=58.26N-F_{2}\\F_{2}=58.26N-25N=33.26N\\\mid F_{2} \mid=\frac{kq_{2}q}{d^2}$

Rewriting for $q_{2}$ :

$q_{2}=\frac{F_{2}d^2}{kq}\\q_{2}=\frac{33.26N(0.34m)^2}{8.99*10^9\frac{Nm^2}{C^2}(8.4*10^{-6}C)}=50.91*10^{-6}C=50.91 \mu C$

8 0

3 years ago

A 5 kgkg sphere having a charge of ++ 8 μCμC is placed on a scale, which measures its weight in newtons. A second sphere having

Mrac [35]

Answer:

F_Balance = 46.6 N ,m' = 4,755 kg

Explanation:

In this exercise, when the sphere is placed on the balance, it indicates the weight of the sphere, when another sphere of opposite charge is placed, they are attracted so that the balance reading decreases, resulting in

∑ F = 0

Fe –W + F_Balance = 0

F_Balance = - Fe + W

The electric force is given by Coulomb's law

Fe = k q₁ q₂ / r₂

The weight is

W = mg

Let's replace

F_Balance = mg - k q₁q₂ / r₂

Let's reduce the magnitudes to the SI system

q₁ = + 8 μC = +8 10⁻⁶ C

q₂ = - 3 μC = - 3 10⁻⁶ C

r = 0.3 m = 0.3 m

Let's calculate

F_Balance = 5 9.8 - 8.99 10⁹ 8 10⁻⁶ 3 10⁻⁶ / (0.3)²

F_Balance = 49 - 2,397

F_Balance = 46.6 N

This is the balance reading, if it is calibrated in kg, it must be divided by the value of the gravity acceleration.

Mass reading is

m' = F_Balance / g

m' = 46.6 /9.8

m' = 4,755 kg

6 0

3 years ago