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3 years ago

Match the element with its description.

Physics

2 answers:

Nutka1998 [239]3 years ago

6 0

Answer:

1.c

2.b

3.d

4.a

Explanation:

There are your answers

zubka84 [21]3 years ago

4 0

Answer:

Sodium---Malleable, soft, and shiny

Silicon---Has properties of both metals and nonmetals

Bromine---Highly reactive gas

Argon---Nonreactive gas

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To test the performance of its tires, a car

yarga [219]

Answer:

0.43

Explanation:

Sum of forces in the y direction:

∑F = ma

N − mg = 0

N = mg

There are friction forces in two directions: centripetal and tangential. The centripetal acceleration is:

ac = v² / r

ac = (35 m/s)² / 564 m

ac = 2.17 m/s²

The total acceleration is:

a = √(ac² + at²)

a = √((2.17 m/s²)² + (3.62 m/s²)²)

a = 4.22 m/s²

Sum of forces:

∑F = ma

Nμ = ma

mgμ = ma

μ = a / g

μ = 4.22 m/s² / 9.8 m/s²

μ = 0.43

5 0

4 years ago

Suppose that you can throw a projectile at a large enough v0 so that it can hit a target a distance R downrange. Given that you

NikAS [45]

Answer:

Theta1 = 12° and theta2 = 168°

The solution procedure can be found in the attachment below.

Explanation:

The Range is the horizontal distance traveled by a projectile. This diatance is given mathematically by Vo cos(theta) t. Where t is the total time of flight of the projectile in air. It is the time taken for the projectile to go from starting point to finish point. This solution assumes the projectile finishes uts motion on the same horizontal level as the starting point and as a result the vertical displacement is zero (no change in height).

In the solution as can be found below, the expression to calculate the range for any launch angle theta was first derived and then the required angles calculated from the equation by substituting the values of the the given quantities.

7 0

3 years ago

The 500-kg cylindrical drum is supported by a metal cable (A-B) and a rigid plate (B-C). If the contact between the drum and all

kotykmax [81]

Explanation:

Beloware attachments containing the complete question and solution.

8 0

4 years ago

Could someone help me to solve this que?

Rashid [163]

Answer:

The speed of the ball B is 6.4 m/s. The direction is 50 degrees counterclockwise.

Explanation:

Assuming the collision is elastic, use the conservation of momentum to solve this problem. The conservation law implies that:

$m\vec v_{A0}+m\vec v_{B0} = m\vec v_{A1}+m\vec v_{B1}$

(the total momentum of the two balls is the same before (index 0) and after (index 1) the collision). Since B is stationary and A and B have the same mass, this simplifies to:

$\vec v_{A0} = \vec v_{A1}+\vec v_{B1}$

and allows us to determine the velocity of ball B after the collision:

$\vec v_{B1} = \vec v_{A0}-\vec v_{A1}$

The above involves vectors. Your problem suggests to use the component method, which I am assuming means solving the above equation separately along the x and y axes. Define x to align with the original line of motion of the ball A before the collision, and y to be perpendicular to x, pointing up:

$v_{B1x} = v_{A0x}-v_{A1x}\\v_{B1y} = v_{A0y}-v_{A1y}$

We just need to compute the x- and y-components of the known velocity of the ball A. Drs. Sine and Cosine come to help here.

$v_{A1x} = |v_{A1}|\cos 40^\circ\\v_{A1y} = |v_{A1}|\sin 40^\circ$

$v_{B1x} = |v_{A0}|\cos 0^\circ-|v_{A1}|\cos 40^\circ=(10.0-7.7\cdot 0.77) \frac{m}{s}\approx 4.1\frac{m}{s}\\v_{B1y} = |v_{A0}|\sin 0^\circ-|v_{A1}|\sin 40^\circ=(0-7.7\cdot 0.64) \frac{m}{s}\approx -4.9\frac{m}{s}$

The speed of the ball B is $|v_{B1}| = \sqrt{4.1^2+(-4.9)^2}\frac{m}{s}\approx 6.4 \frac{m}{s}$ . The direction (angle from horizontal) is $\beta = \arcsin (-\frac{4.9}{6.4})\approx -50^\circ$ , i.e., 50 degrees counterclockwise.