How many teeth shows signs of decay?

A traffic light of weight 100 N is supported by two ropes that make 30-degree angle with the horizontal. What is the vertical co

alina1380 [7]

Answer:

$T_{vertical} = 50 N$

Explanation:

given,

traffic light weight = 100 N

angle at which the rope is supported = 30°

vertical component of force = ?

$2 T sin \theta= W$

$2 T sin 30^0= 100$

$T = \dfrac{100}{2 sin 30^0}$

$T = 100\ N$

$T_{vertical} = T sin 30^0$

$T_{vertical} = 100\times \dfrac{1}{2}$

$T_{vertical} = 50 N$

5 0

4 years ago

A 4.0 kg circular disk slides in the x- direction on a frictionless horizontal surface with a speed of 5.0 m/s as shown in the a

finlep [7]

Solution :

Let $m_1=m_2=4$ kg

$u_1 = 5$ m/s

Let $v_1$ and $v_2$ are the speeds of the disk $m_1$ and $m_2$ after the collision.

So applying conservation of momentum in the y-direction,

$0=m_1 .v_1_y -m_2 .v_2_y$

$v_1_y = v_2_y$

$v_1 . \sin 60=v_2. \sin 30$

$v_2 = v_1 \times \frac{\sin 60}{\sin 30}$

$v_2=1.732 \times v_1$

Therefore, the disk 2 have greater velocity and hence more kinetic energy after the collision.

Now applying conservation of momentum in the x-direction,

$m_1.u_1=m_1.v_1_x+m_2.v_2_x$

$u_1=v_1_x+v_2_x$

$5=v_1. \cos 60 + v_2 . \cos 30$

$5=v_1. \cos 60 + 1.732 \times v_1 \cos 30$

$v_1 = 2.50$ m/s

So, $v_2 = 1.732 \times 2.5$

= 4.33 m/s

Therefore, speed of the disk 2 after collision is 4.33 m/s

5 0

3 years ago

11. The primary job of the cell wall is to

hodyreva [135]

11. protect the cell and keep its shape.

12. chloroplast

3 0

3 years ago

Read 2 more answers

A car rounds a curve. The radius of curvature of the road is R, the banking angle with respect to the horizontal is θ and the co

exis [7]

Answer:

v = √[gR (sin θ - μcos θ)]

Explanation:

The free body diagram for the car is presented in the attached image to this answer.

The forces acting on the car include the weight of the car, the normal reaction of the plane on the car, the frictional force on the car and the net force on the car which is the centripetal force on the car keeping it in circular motion without slipping.

Resolving the weight into the axis parallel and perpendicular to the inclined plane,

N = mg cos θ

And the component parallel to the inclined plane that slides the body down the plane at rest = mg sin θ

Frictional force = Fr = μN = μmg cos θ

Centripetal force responsible for keeping the car in circular motion = (mv²/R)

So, a force balance in the plane parallel to the inclined plane shows that

Centripetal force = (mg sin θ - Fr) (since the car slides down the plane at rest, (mg sin θ) is greater than the frictional force)

(mv²/R) = (mg sin θ - μmg cos θ)

v² = R(g sin θ - μg cos θ)

v² = gR (sin θ - μcos θ)

v = √[gR (sin θ - μcos θ)]

Hope this Helps!!!

5 0

3 years ago

Given vectors D (3.00 m, 315 degrees wrt x-axis) and E (4.50 m, 53.0 degrees wrt x-axis), find the resultant R= D + E. (a) Write

fomenos

Answer:

(a) $\vec{R}= 4.83\ m\ \hat{i}+1.47\ m\ \hat{j}$

(b) (5.05 m, 16.93 degrees wrt x-axis)

Explanation:

Given:

$\vec{D}$ = (3.00 m, 315 degrees wrt x-axis)
$\vec{E}$ = (4.50 m, 53.0 degrees wrt x-axis)

Let us first fond out vector D and E in their rectangular form.

$\vec{D} = (3\cos 315^\circ\ \hat{i}+3\sin 315^\circ\ \hat{j})\ m\\\Rightarrow \vec{D} = (2.12\ \hat{i}-2.12\ \hat{j})\ m\\$

Similarly,

$\vec{E} = (4.5\cos 53^\circ\ \hat{i}+4.5\sin 53^\circ\ \hat{j})\ m\\\Rightarrow \vec{D} = (2.71\ \hat{i}+3.59\ \hat{j})\ m\\\because \vec{R}=\vec{D}+\vec{E}\\\therefore \vec{R} = (2.12\ \hat{i}-2.12\ \hat{j})\ m+(2.71\ \hat{i}+3.59\ \hat{j})\ m\\\Rightarrow \vec{R} = (4.83\ \hat{i}+1.47\ \hat{j})\ m$

Part (a):

We can write the resultant vector R as below:

$\vec{R} = (4.83\ \hat{i}+1.47\ \hat{j})\ m$

Part (b):

$Magnitude\ of\ resultant = \sqrt{4.83^2+1.47^2}\ m = 5.05\ m\\\textrm{Direction in angle with the x-axis} = \theta = \tan^{-1}(\dfrac{1.47}{4.83})= 16.93^\circ$

Since both the components of the resultant lie on the positive x and y axes. So, the resultant makes an acute angle with the positive x-axis.

So, R = (5.05 m, 16.93 degrees wrt x-axis)

3 0

3 years ago

How many teeth shows signs of decay?​

How many teeth shows signs of decay?