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uranmaximum [27]

3 years ago

5

Part A: Explain why x = 5 makes 4x − 1 ≤ 19 true but not 4x − 1 < 19. (5 points) Part B: What value from the set {6, 7, 8, 9,

10} makes the equation 5x + 2 = 47 true? Show your work. (5 points)

1 answer:

Elis [28]3 years ago

5 0

Answer:

A: ≤ means less than OR equal to. < only means less than

B: 9

Explanation:

A: Because it would equal 19, and 19 is EQUAL than 19. 4(5) - 1 would equal 19, which is equal to 19, and not less than. ≤ means less than or equal to. < means less than. So its not true.

B: 47 - 2, 45. Then 5 x 9 equals 45. So 5 x 9 equals 45, then add 2 would equal 47.

Hope this helps <3

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a ball kicked with a velocity of 8m/s at an angle of 30 degree to horizontal. calculate the time of flight of the ball. (g=10ms^

posledela

Answer:

Approximately $0.8\; \rm s$ (assuming that air resistance is negligible.)

Explanation:

Let $v_0$ denote the initial velocity of this ball. Let $\theta$ denote the angle of elevation of that velocity.

The initial velocity of this ball could be decomposed into two parts:

Initial vertical velocity: $v_0(\text{vertical}) = v_0 \cdot \sin(\theta)$ .
Initial horizontal velocity: $v_0(\text{vertical}) = v_0 \cdot \cos(\theta)$ .

If air resistance on this ball is negligible, $v_0(\text{vertical})$ alone would be sufficient for finding the time of flight of this ball.

Calculate $v_0(\text{vertical})$ given that $v_0 = 8 \; \rm m \cdot s^{-1}$ and $\theta = 30^\circ$ :

$\begin{aligned}& v_0(\text{vertical}) \\ &= v_0 \cdot \sin(\theta) \\ &= \left(8 \; \rm m \cdot s^{-1} \right) \cdot \sin\left(30^{\circ}\right) \\ &= 4\;\rm m \cdot s^{-1} \end{aligned}$ .

Assume that air resistance on this ball is zero. Right before the ball hits the ground, the vertical velocity of this ball would be exactly the opposite of the value when the ball was launched.

Since $v_0(\text{vertical}) = 4\; \rm m \cdot s^{-1}$ , the vertical velocity of this ball right before landing would be $v_1(\text{vertical}) = -4\; \rm m \cdot s^{-1}$ .

Calculate the change to the vertical velocity of this ball:

$\begin{aligned}& \Delta v(\text{vertical}) \\ & = v_1(\text{vertical}) - v_0(\text{vertical}) \\ &= -8\; \rm m \cdot s^{-1}\end{aligned}$ .

In other words, the vertical velocity of this ball should have change by $8\; \rm m \cdot s^{-1}$ during the entire flight (from the launch to the landing.)

The question states that the gravitational field strength on this ball is $g = 10\; \rm m \cdot s^{-2}$ . In other words, the (vertical) downward gravitational pull on this ball could change the vertical velocity of the ball by $10\; \rm m\cdot s^{-1}$ each second. What fraction of a second would it take to change the vertical velocity of this ball by $8\; \rm m \cdot s^{-1}$ ?

$\begin{aligned}t &= \frac{\Delta v(\text{initial})}{g} \\ &= \frac{8\; \rm m \cdot s^{-1}}{10\; \rm m \cdot s^{-2}} = 0.8\; \rm s\end{aligned}$ .

In other words, it would take $0.8\; \rm s$ to change the velocity of this ball from the initial velocity at launch to the final velocity at landing. Therefore, the time of the flight of this ball would be $0.8\; \rm s\!$ .

5 0

3 years ago

The block in the diagram below is AT REST. However, the tension in the cable is not the only thing holding the block back. Stati

Vedmedyk [2.9K]

Answer:

The tension in the rope is 229.37 N.

Explanation:

Given:

Mass of the block is, $m=33.2\ kg$

Coefficient of static friction is, $\mu = 0.214$

Angle of inclination is, $\theta = 31.5$ °

Draw a free body diagram of the block.

From the free body diagram, consider the forces in the vertical direction perpendicular to inclined plane.

Forces acting are $mg\cos \theta$ and normal $N$ . Now, there is no motion in the direction perpendicular to the inclined plane. So,

$N=mg\cos \theta\\N=(33.2)(9.8)\cos (31.5)\\N=277.415\ N$

Consider the direction along the inclined plane.

The forces acting along the plane are $mg\sin \theta$ and frictional force, $f$ , down the plane and tension, $T$ , up the plane.

Now, as the block is at rest, so net force along the plane is also zero.

$T=mg\sin \theta+f\\T=mg\sin \theta +\mu N\\T= (33.2)(9.8)(\sin (31.5)+(0.214\times 277.415)\\T= 170+59.37\\T=229.37\ N$

Therefore, the tension in the rope is 229.37 N.

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