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2 years ago

Can someone please help me with part a and d

Physics

2 answers:

Temka [501]2 years ago

7 0

Answer: question D is trough

Explanation: I learned this already

Semenov [28]2 years ago

4 0

Answer:

'a' is the crest while 'd' is the trough

Explanation:

'a' is the crest of the wave and it is defined as the maximum displacement of a wave from its initial position. The value of crest is a positive amplitude.

'd' is known as the trough and is the defined as the minimum displacement of an object from its initial position to produce a wave motion.

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a swimmer can swim in still water at a speed of 9.50 m/s. he intends to swim directly across the river that has a downstream cur

Doss [256]

Refer to the diagram shown below.

Still-water speed = 9.5 m/s
River speed = 3.75 m/s down stream.

The velocity of the swimmer relative to the bank is the vector sum of his still-water speed and the speed of the river.

The velocity relative to the bank is
V = √(9.5² + 3.75²) = 10.21 m/s

The downstream angle is
θ = tan⁻¹ 3.75/9.5 = 21.5°

Answer: 10.2 m/s at 21.5° downstream.

7 0

3 years ago

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A force in the +x-direction with magnitude ????(x) = 18.0 N − (0.530 N/m)x is applied to a 6.00 kg box that is sitting on the ho

fiasKO [112]

Answer:

$v_f=8.17\frac{m}{s}$

Explanation:

First, we calculate the work done by this force after the box traveled 14 m, which is given by:

$W=\int\limits^{x_f}_{x_0} {F(x)} \, dx \\W=\int\limits^{14}_{0} ({18N-0.530\frac{N}{m}x}) \, dx\\W=[(18N)x-(0.530\frac{N}{m})\frac{x^2}{2}]^{14}_{0}\\W=(18N)14m-(0.530\frac{N}{m})\frac{(14m)^2}{2}-(18N)0+(0.530\frac{N}{m})\frac{0^2}{2}\\W=252N\cdot m-52N\cdot m\\W=200N\cdot m$

Since we have a frictionless surface, according to the the work–energy principle, the work done by all forces acting on a particle equals the change in the kinetic energy of the particle, that is:

$W=\Delta K\\W=K_f-K_i\\W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}$

The box is initially at rest, so $v_i=0$ . Solving for $v_f$ :

$v_f=\sqrt{\frac{2W}{m}}\\v_f=\sqrt{\frac{2(200N\cdot m)}{6kg}}\\v_f=\sqrt{66.67\frac{m^2}{s^2}}\\v_f=8.17\frac{m}{s}$

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2 years ago

Predict the precipitate when the double replacement reaction occurs:<br> Ca(NO3)2(aq) + Na2CO3(aq) →

taurus [48]

Ca(NO3)2(aq) + Na2CO3(aq) → 2NaNO3 + CaCO3⬇. NaNO3 is solution so CaCO3 is the precipitate formed.

8 0

3 years ago

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In a football game, running back is at the 10 yard line and running up the field towards the 50 yard line, and runs for 3 second

lakkis [162]

If he keeps that pace he will be at the 34 yard line

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2 years ago

A 5 kg ball takes 13.3 seconds for one revolution around the circle. What's the magnitude of the angular velocity of this motion

Tcecarenko [31]

Answer: 0.47 rad/sec

Explanation:

By definition, the angular velocity is the rate of change of the angle traveled with time, so we can state the following:

ω = ∆θ/ ∆t

Now, we are told that in 13.3 sec, the ball completes one revolution around the circle, which means that, by definition of angle, it has rotated 2 π rad (an arc of 2πr over the radius r), so we can find ω as follows:

ω = 2 π / 13.3 rad/sec = 0.47 rad/sec

6 0

3 years ago

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