All categories

3 years ago

Calculate the parallel component of the weight of the object on the inclined plane below.

Physics

1 answer:

const2013 [10]3 years ago

8 0

Answer:

4. Wx = 42.47[N]

Explanation:

In order to solve this problem, we must first make a free body diagram, of the forces acting on the body in the inclined plane.

As we can see in the attached image, the component of the weight parallel to the surface of the inclined plane can be obtained by the sine of the angle.

$W_{x}=5*9.81*sin(60)\\W_{x}=42.47[N]$

You might be interested in

A stationary 90 kg baseball pitcher throws a 0.15 kg baseball forward

Marina CMI [18]

Answer: momentum = 6kgm/s

Explanation:

given that the baseball pitcher is at stationary position, his velocity will be equal to zero. If velocity is zero, his linear momentum will therefore equal to zero.

Linear momentum is the product of mass and velocity. Given that the baseball has

Mass M = 0.15 kg

Velocity V = 40 m/s

Momentum = MV

Momentum = 0.15 × 40 = 6 kgm/s

4 0

2 years ago

The charges are in a uniform electric field whose direction makes an angle 36.8 ∘ with the line connecting the charges. What is

astraxan [27]

Complete question:

Point charges q1=- 4.10nC and q2=+ 4.10nC are separated by a distance of 3.60mm , forming an electric dipole. The charges are in a uniform electric field whose direction makes an angle 36.8 ∘ with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.30×10−9 N⋅m ? Express your answer in newtons per coulomb to three significant figures.

Answer:

The magnitude of this field is 826 N/C

Explanation:

Given;

The torque exerted on the dipole, T = 7.3 x 10⁻⁹ N.m

PEsinθ = T

where;

E is the magnitude of the electric field

P is the dipole moment

First, we determine the magnitude dipole moment;

Magnitude of dipole moment = q*r

P = 4.1 x 10⁻⁹ x 3.6 x 10⁻³ = 1.476 x 10⁻¹¹ C.m

Finally, we determine the magnitude of this field;

$E = \frac{T}{P*sin(\theta)}= \frac{7.3 X 10^{-9}}{1.476X10^{-11}*sin(36.8)}\\\\E = 825.6 N/C$

E = 826 N/C (in three significant figures)

Therefore, the magnitude of this field is 826 N/C

6 0

3 years ago

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

Distance of the center from each corners $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

Force by each of the charges at the remaining corners:

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

Now, net electric field in the vertical direction:

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

Now, net electric field in the horizontal direction:

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

8 0

3 years ago

Institutions that anticipate and recognize significant external forces and modify their strategies and operations will prosper a

vesna_86 [32]

Answer:True

Explanation: External factors are factors outside of an organisation or an institution that has the capacity of either adversely or positively affect the institution or the Organisation.

For an institution to prosper and perform optimally in an economy.or a country, such Institution must put into cognisance the possible external threats that are capable of affecting it, when Organisations or Institution put strategies in place to control or mitigate such externalities,the institution or Organisation will sure Prosper.

5 0

3 years ago

15. In the diagram shown, the emf of the secondary winding is______ the emf of the primary winding.

sp2606 [1]

Answer:

Greater than

Explanation:

3 0

2 years ago