What would be the distance moved if we had a 70 n force and work done is 8j

What is the area of the parallelogram?<br>square units

Damm [24]

The area is the length times the height so A= bh ...?

3 0

3 years ago

Which phase of matter contains particles that split into ions and electrons?

Lemur [1.5K]

Answer:

Plasma

Explanation:

7 0

3 years ago

If a man does 4500 J of work from a high rise building of 10 m, what is his mass?

GalinKa [24]

Explanation:

The potential energy (Ep) = 4500 J

Ep = m. g. h

=> m = Ep/(g.h)

m = 4500/(10×10)

= 4500/100

= 45

so, the mass = 45 kgs

4 0

3 years ago

Read 2 more answers

Consider a semicircular ring of radius R. Its linear mass density varies as lambda =lambda not sin theta. Locate its centre of m

bearhunter [10]

Answer:

(0, πR/4)

Explanation:

The linear mass density (mass per length) is λ = λ₀ sin θ.

A short segment of arc length is ds = R dθ.

The mass of this short length is:

dm = λ ds

dm = (λ₀ sin θ) (R dθ)

dm = R λ₀ sin θ dθ

The x coordinate of the center of mass is:

X = ∫ x dm / ∫ dm

X = ∫₀ᵖ (R cos θ) (R λ₀ sin θ dθ) / ∫₀ᵖ R λ₀ sin θ dθ

X = R ∫₀ᵖ sin θ cos θ dθ / ∫₀ᵖ sin θ dθ

X = R ∫₀ᵖ ½ sin 2θ dθ / ∫₀ᵖ sin θ dθ

X = ¼R ∫₀ᵖ 2 sin 2θ dθ / ∫₀ᵖ sin θ dθ

X = ¼R (-cos 2θ)|₀ᵖ / (-cos θ)|₀ᵖ

X = ¼R (-cos 2π − (-cos 0)) / (-cos π − (-cos 0))

X = ¼R (-1 + 1) / (1 + 1)

X = 0

The y coordinate of the center of mass is:

Y = ∫ y dm / ∫ dm

Y = ∫₀ᵖ (R sin θ) (R λ₀ sin θ dθ) / ∫₀ᵖ R λ₀ sin θ dθ

Y = R ∫₀ᵖ sin² θ dθ / ∫₀ᵖ sin θ dθ

Y = R ∫₀ᵖ ½ (1 − cos 2θ) dθ / ∫₀ᵖ sin θ dθ

Y = ½R ∫₀ᵖ (1 − cos 2θ) dθ / ∫₀ᵖ sin θ dθ

Y = ½R (θ − ½ sin 2θ)|₀ᵖ / (-cos θ)|₀ᵖ

Y = ½R [(π − ½ sin 2π) − (0 − ½ sin 0)] / (-cos π − (-cos 0))

Y = ½R (π − 0) / (1 + 1)

Y = ¼πR

4 0

3 years ago

Two blocks connected by a light string are being pulled across a frictionless horizontal tabletop by a hanging 16.2-N weight (bl

Artemon [7]

Newton's second law allows us to find the results for the string tensions are:

T₁ = 6.7 N

T₂ = 16.54 N

Newton's second law gives a relationship between force, mass and acceleration of bodies

∑ F = ma

Where the bold letters indicate vectors, F is the force, m the mass and the acceleration.

Free-body diagrams are representations of the forces applied to bodies without the details of them.

The reference system is a coordinate system with respect to which the forces decompose, in this case the x-axis is parallel to the plane and the positive direction in the direction of movement, the y-axis is perpendicular to the plane.

In the attachment we see a free-body diagram of the three-block system.

Let's apply Newton's second law to each body.

Block C

Y-axis

$W_c -T_2 = m_c a$

Block A

X axis

$T_2 - T_1 - W_a_x = m_a a$

Y axis

$N_a - W_a_y = 0$

Block B

X axis

$T_1 - W_b_x = m_b a$

Y axis

$N_b - W_b_y =0$

Let's use trigonometry to find the components of the weight.

Block A

cos θ = $\frac{W_a_y}{W_a}$

sin θ = $\frac{W_a_x}{W_a}$

$W_a_y = W_a cos \theta$

$W_a_x= W_a sin \theta$

Block B

cos θ = $\frac{W_b_y}{W_b}$

sin θ = $\frac{W_b_x}{W_b}$

$W_b_y = W_b cos \theta \\W_b_x = W_b sin \theta$

Let's write our system of equations.

$W_c - T_2 = m_c a \\ T_2 - T_1 - W_a_x = m_a a \\T_1 - W_b_x = m_b a$

Let's find the acceleration of the bodies, adding the equations.

$W_c - W_a_x - W_b_x = ( m_a+m_b+m_c) a\\$

The weight is

W = mg

Let's substitute

$(m_c - m_a -m_b ) g \ sin \theta = ( m_c+m_a+m_b) \ a \\a= \frac{ m_c-m_a-m_b }{ m_a+m_b+m_c} \ g sin \theta$

Indicate ma mass of the block a ma = 1.00 kg, the mass of the block b mb = 2.2 kg and the weight of the block c Wc = 16.2 N, let's find the mass of block c.

m_c = Wc / g

m_c = 16.2 / 9.8

m_c = 1.65 kg

we substitute the values

$a= \frac{1.65 -2.20 -1.00}{1.65+2.20+1.00} \ 9.8 \ sin \theta \\a= -0.3096 sin \theta$

The negative sign indicates that the system is descending, to be able to give a specified value an angle is needed, they assume that the angle of the ramp is 45º

a = - 0.3196 sin 45

a = -0.226 m / s

Taking the acceleration we are going to look for the tensions.

From the equation of block C

$W_c - T_2 = m_c a \\T_2 = m_c ( g-a)\\T_2 = 1.65 ( 9.8 + 0.226)$