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3 years ago

Which of the following statements is true?

Physics

1 answer:

fenix001 [56]3 years ago

4 0

Answer:Light (electromagnetic) energy does not require a medium through which to travel.

Explanation:

Electromagnetic waves doesnt need to medium for travell

Ex ; Ultra violet rays , X rays

Sound waves needs to medium to travell but electromagnetic waves doesnt

Electromagnetic waves are constructed due to perpendicular electric field and magnetic field

Application of electromagnetic waves

Visible rays helps to see anything

IR rays is used in medical industry for cancer treatment

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"When Baby Ian is held horizontally on his back, he sometimes makes an embracing motion by arching his back, extending his legs,

Burka [1]

Answer:

Moro reflex

Explanation:

The baby has a series of reflexes in its first months of life and they are very important for the good development of the little ones. One of them is the Moro Reflex. In this reflection, the baby spreads his arms wide, stretches his legs and extends his neck. Moro's reflex begins from birth and continues until about four months of the baby. It occurs when the baby abruptly shifts position or falls backwards or feels in an insecure position, at which time the baby makes a hug movement by arching his back, extending his legs, throwing his arms out and then bringing his arms. towards the body.

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3 years ago

In a thermostat, what property of the bimetallic coil allows it to contract and expand? The two metals absorb different amounts

Reika [66]

Hello!

In a thermostat, the property of the bimetallic coil that allows it to contract and expand is that The two metals absorb different amounts of thermal energy.

This bimetallic coil is used to transform thermal energy into mechanical movement. Two metals with different thermal expansivity are joined together parallelly and the changes of temperature cause bending in different directions depending on if the temperature is rising or descending.

The differences in the thermal energy absorption of the two metals are the basis for the mechanism of this device.

3 0

3 years ago

Read 2 more answers

Long Jump: inital center of mass height of 1.08 m, final center of mass height of 0.42 m, projection velocity of 8.7 m/s, projec

sammy [17]

Answer:

1) The maximum jump height is reached at A. $0.337s$

2) The maximum center of mass height off of the ground is B. $1.64m$

3) The time of flight is C. $0.834s$

4) The distance of jump is B. $7.49m$

Explanation:

First of all we need to decompose velocity in its rectangular components, so

$v_{xi}=8.7m/s(cos 22.3\°)=8.05m/s= constant\\v_{yi}=8.7m/s(sin 22.3\°)=3.3m/s$

1) We use, $v_{fy}=v_{iy}-gt$ , as we clear it for $t$ and using the fact that $v_{fy}=0$ at max height, we obtain $t=\frac{v_{iy}}{g} =\frac{3.3m/s}{9,8m/s^{2}} =0.337s$

2) We can use the formula $y_{max}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ for $t=0.337s$ , so

$y_{max}=1.08m+(3.3m/s)(0.337s)-\frac{(9.8m/s^{2})(0.337)^{2}}{2}=1.64m$

3) We can use the formula $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find total time of fligth, so $0.42=1.08+3.3t-\frac{(9.8)t^{2}}{2}\\0=-4.9t^{2}+3.3t+0.66$ , as it is a second-grade polynomial, we find that its positive root is $t=0.834s$

4) Finally, we use $x=v_{x}t=8.05m/s(0.834s)=6.71m$ , as it has an additional displacement of $0.77m$ due the leg extension we obtain,

$x=6.71m+0.77m=7.48m$ , aprox $x=7.49m$

5 0

3 years ago

3.00 textbook rests on a frictionless, horizontal tabletop surface. A cord attached to the book passes over a pulley whose diame

sammy [17]

Answer:

a1 = 3.56 m/s²

Explanation:

We are given;

Mass of book on horizontal surface; m1 = 3 kg

Mass of hanging book; m2 = 4 kg

Diameter of pulley; D = 0.15 m

Radius of pulley; r = D/2 = 0.15/2 = 0.075 m

Change in displacement; Δx = Δy = 1 m

Time; t = 0.75

I've drawn a free body diagram to depict this question.

Since we want to find the tension of the cord on 3.00 kg book, it means we are looking for T1 as depicted in the FBD attached. T1 is calculated from taking moments about the x-axis to give;

ΣF_x = T1 = m1 × a1

a1 is acceleration and can be calculated from Newton's 2nd equation of motion.

s = ut + ½at²

our s is now Δx and a1 is a.

Thus;

Δx = ut + ½a1(t²)

u is initial velocity and equal to zero because the 3 kg book was at rest initially.

Thus, plugging in the relevant values;

1 = 0 + ½a1(0.75²)

Multiply through by 2;

2 = 0.75²a1

a1 = 2/0.75²

a1 = 3.56 m/s²

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2 years ago

Electric Circuits Homework

nirvana33 [79]

<h2>Sorry, But I don't know!!</h2>

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2 years ago