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mr Goodwill [35]
2 years ago
10

Bill drives west at 20\, \dfrac{\text m}{\text s}20 s m ​ 20, start fraction, start text, m, end text, divided by, start text, s

, end text, end fraction, then slows to a stop at the red light. A graph of his velocity over time is shown below, where eastward is the positive velocity direction.
Physics
2 answers:
lara [203]2 years ago
7 0

Answer:

This

Explanation:

Khan academy answer

vovangra [49]2 years ago
4 0

Answer:

Positive Horizontal line, constant

Explanation:

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Which of the following statements is TRUE?
Yuri [45]
Hello there..

<span>Which of the following statements is TRUE?

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A magnetic field is passing through a loop of wire whose area is 0.020 m2. The direction of the magnetic field is parallel to th
Evgen [1.6K]

Answer:

a) E = 0.0048 Volts

b)  dA/dt = - 0.002285 m²/s

Explanation:

Given:

Area, A = 0.020 m²

Rate of change of magnetic field, dB/dt = 0.24 T/s

a) The magnitude of the emf induced (E) is given as:

  E= A × (dB/dt)

  on substituting the values in the above equation, we get

E = (0.020 m²) × (0.24 T/s)

or

E = 0.0048 Volts

b) Now, The induced emf when both the area and the magnetic field is varying

we have

E = B(dA/dt) + A(dB/dt)

Now, for the given case induced emf is zero i.e E = 0 and magnetic field B = 2.1 T

thus,

0 = (2.1 T)(dA/dt) + (0.020 m2)(0.24 T/s)

dA/dt = - 0.002285 m²/s

Hence, the area should be decreased at the rate of 0.002285 m²/s

3 0
3 years ago
In 2007, michael carter (u.s.) set a world record in the shot put with a throw of 24.77 m. what was the initial speed of the sho
Serga [27]

The initial speed of the shot is 15.02 m/s.

The Shot put is released at a height y<em> </em>from the ground with a speed u. It is released at an angle θ to the horizontal. In a time t, the shot put travels a distance <em>R</em> horizontally.

Pl refer to the attached diagram.

Resolve the velocity u into horizontal and vertical components, u ₓ=ucosθ and uy=u sinθ. The horizontal component remains constant in the absence of air resistance, while the vertical component varies due to the action of the gravitational force.

Write an expression for R.

R=u_xt=(ucos \theta)t

Therefore,

t=\frac{R}{ucos\theta} .......(1)

In the time t, the net displacement of the shotput is y in the downward direction.

Use the equation of motion,

y=u_yt-\frac{1}{2}gt^2=(usin\theta) t-\frac{1}{2}gt^2

Substitute the value of t from equation (1).

y=(ucos\theta)(\frac{R}{ucos\theta} )-\frac{1}{2} g(\frac{R}{ucos\theta} )^2\\ =Rtan\theta-(\frac{gR^2}{2u^2cos^2\theta} )

Substitute -2.10 m for y, 24.77 m for R and 38.0° for θ and solve for u.

y=Rtan\theta-(\frac{gR^2}{2u^2cos^2\theta} )\\ (-2.10m)=(24.77 m)(tan38.0^o)-\frac{(9.8 m/s^2)(24.77m)^2}{2u^2(cos38.0^o)^2} \\ u^2=225.71(m/s)^2\\ u=15.02m/s

The shot put was thrown with a speed 15.02 m/s.




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Which of the following is a physical change
topjm [15]

Answer: A physical change can be a change in size, shape, or color.

Explanation:

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I am Lyosha [343]
His acceleration would be 3.34 m/s 
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