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Natasha2012 [34]
2 years ago
7

Which of the following is an example of a chemical change?

Physics
1 answer:
tankabanditka [31]2 years ago
3 0

b. burning paper

hope it helps

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The average coefficient of linear expansion of copper is 1.7 10-5 (°c)−1. the statue of liberty is 93 m tall on a summer morning
sammy [17]

Let the rise in temperature be 5^0C

The expansion in length due to change in temperature is given by the expression lαΔt , where l is the length, α is the  coefficient of linear expansion, Δt is the change in temperature.

Here l = 93 m, α = 1.7*10^{-5}  ^0C^{-1}, and Δt = 5^0C

So expansion in length = 93*1.7*10^{-5}*5 = 0.007905 m = 0.79*10^{-3}m

So order of magnitude in change in length = -3


3 0
3 years ago
Read 2 more answers
A cart loaded with bricks has a total mass of 9.13 kg and is pulled at constant speed by a rope. The rope is inclined at 24.7 ◦
blagie [28]

Answer:

W = 0.63 KJ

Explanation:

Work (W) is defined as the point product of force (F) by the distance (d)the body travels due to this force.  

W= F*d *cosα Formula (1)  

F : force (N)

d : displacement (m)

α : angle between force and displacement

Newton's second law:

∑F = m*a Formula (2)  

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

We define the x-axis in the direction parallel to the movement of the cart on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the cart

W: Weight of the cart : In vertical direction

FN : Normal force : perpendicular to the floor

f : Friction force: parallel to the floor

T : tension Force,  inclined at  θ=24.7° above the horizontal

Calculated of the W

W= m*g

W= 9.13 kg* 9.8 m/s² = 89.47 N

x-y components o the  tension force (T)

Tx = Tcosθ = T*cos 24.7° (N)

Ty = Tsin θ = T*sin 24.7°  (N)

Calculated of the FN  

We apply the formula (2)  

∑Fy = m*ay ay = 0  

FN +Ty- W = 0  

FN = W-Ty  

FN =  89.47-T*sin 24.7°

Calculated of the friction force (f)

f = μk*FN

f =(0.597)*(  89.47-T*sin 24.7° )

f= 53.41-0.249T

Calculated of the tension force of the rope (f)

We apply the formula (2) :

∑Fx = m*ax  ,  ax= 0 ,because the speed of the cart  is constant

Tx - f = 0

T*cos 24.7°-( 53.41 - 0.249T )= 0

T*cos 24.7° + 0.249T = 53.41

(1.1575)T = 53.41

T= (53.41) / (1.1575)

T= 46.14 N

Work done on the cart by the rope

We apply the formula (1)

W=T*d *cosα

W= (46.14 N)*(15.1 m) *(cos24.7)

W = 632.97 (N*m) = 632.97 (J)

W = 0.63 KJ

6 0
3 years ago
Which of the following statements is always true?
Gwar [14]

Answer: B

Explanation:

An unbalanced force (net force) acting on an object changes its speed and/or direction of motion. An unbalanced force is an unopposed force that causes a change in motion.

7 0
3 years ago
Can someone solve this problem and explain to me how you got it​
evablogger [386]

Answer:

question5: F=74312.5N

question6: charge at the end of antenna=0.37N

Explanation:

Coulomb's law: the magnitude of the force of attraction or repulsion due to two charges is proportional to the product of the magnitude of the charges and inversely proportional to the square of distance between the charges.

⇒F\alpha\frac{q1*q2}{r^{2}}

∴F=k\frac{q1*q2}{r^{2}}

where F is the force of attraction or repulsion

k is Coulumb's constant=9*10^{9}Nm^{2}C^{-2}

q1 and q2 are the magnitude of the charges

r is the distance between two charges

The force between the two charges is attractive if they are of different polarity

The force between the two charges is repulsive if they are of same polarity

Question5:

Given: q1=0.041 C, q2=0.029 C, r=12 m

therefore by Coulumb's law,

F=9*10^{9}*\frac{0.041*0.029}{12^{2}}

F=74312.5N

Question6:

Given: q1=3*10^{-18}C, r=5 m, F=4*10^{-11}N

therefore by Coulumb's law,

4*10^{-11}=9*10^{9}*\frac{3*10^{-18}*q2}{5^{2}}

⇒q2=\frac{4*10^{-11}*25}{9*10^{9}*3*10^{-18}} \\=0.37C

4 0
3 years ago
Illustrate a body of mass 5.0kg pulled by horizontal force F . if the body accelerates 2.0ms² and experience a frictional force
Natasha2012 [34]

Explanation:

a. Net force is mass times acceleration (Newton's second law).

∑F = ma

∑F = (5.0 kg) (2.0 m/s²)

∑F = 10 N

b. The net force is the sum of the individual forces.

10 N = F − 5 N

F = 15 N

c. Friction force here is mgμ.

mgμ = 5 N

(5.0 kg) (10 m/s) μ = 5 N

μ = 0.1

3 0
3 years ago
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