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AfilCa [17]
3 years ago
12

Here are three pairs of initial and final positions, respectively, along an x-axis. Which pair gives a

Physics
1 answer:
Sladkaya [172]3 years ago
6 0

Answer:

(b)-3 m,-7 m

Explanation:

Displacement is a vector quantity that connects the initial position of an object to its final position after a certain motion.

Mathematically, displacement can be written as:

d=x_f - x_i

where

x_f is the final position

x_i is the initial position

The direction of this vector is from the initial position to the final position.

In order to calculate the displacement of an object, therefore, it is necessary to correctly take into account the sign of each motion, which gives the direction of each motion.

Let's analyze the three cases:

(a) -3 m, +5 m;

In this case, the object moves 3 m in the negative direction first, and then 5 meters in the positive direction; so the net displacement is

d = -3 + 5 = +2 m

(b)-3 m,-7 m;

In this case, the object moves 3 m in the negative direction first, and then 3 meters in the negative direction; so the net displacement is

d = -3 - 7 = -10 m

(c) 7 m, -3 m?

In this case, the object moves 7 m in the positive direction first, and then 3 meters in the negative direction; so the net displacement is

d = +7 + (-3) = +4 m

So, the only situation in which we have a net negative displacement is situation b).

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The Eiffel tower in Paris is 300 meters tall on a cold day (T = -24 degrees Celsius), what is its height on a hot day when the t
Varvara68 [4.7K]

Answer:

Length of Eiffel tower, when the temperature is 35 degrees = 300.21 m

Explanation:

Thermal expansion is given by the expression

\Delta L=L\alpha \Delta T \\

Here length of Eiffel tower, L = 300 m

Coefficient of thermal expansion, α = 0.000012 per degree Celsius

Change in temperature, = 35 - (-24) = 59degrees Celsius

Substituting

\Delta L=L\alpha \Delta T= 300\times 0.000012\times 59=0.2124m \\

Length of Eiffel tower, when the temperature is 35 degrees = 300 + 0.2124 = 300.21 m

3 0
3 years ago
A car is traveling at 39.7 mi/h on a horizontal highway. The acceleration of gravity is 9.8 m/s 2 . If the coefficient of fricti
777dan777 [17]

Answer:

The minimum distance in which the car will stop is

x=167.38m

Explanation:

39.7\frac{mi}{h}*\frac{1km}{0.621371mi}*\frac{1000m}{1km}*\frac{1h}{3600s}=17.747\frac{m}{s}

∑F=m*a

∑F=u*m*g

The force of friction is the same value but in different direction of the force moving the car so it can stop so

F=m*a\\a=\frac{F}{m}\\a=\frac{u*m*g}{m}\\a=u*g\\a=0.096*-9.8\frac{m}{s^{2} }

a=-0.9408 \frac{m}{s^{2}}

v_{f}^{2}=v_{o}^{2}+2*a*(x_{f}-x_{o})\\v_{f}=0 \\x_{o}=0\\0=v_{o}^{2}+2*a*x_{f}\\x_{f}=\frac{v_{o}^{2}}{2*a} \\x_{f}=\frac{(-17.747\frac{m}{s})^{2}}{2*(-0.9408)} \\x_{f}=167.38m

4 0
3 years ago
Explain the relationship between resistance and the velocity of shortening.
butalik [34]
The velocity of shortening refers to the speed of the contraction from the muscle shortening while lifting a load. The relationship between the resistance and velocity of shortening is inverse. The greater the resistance, the shorter the velocity of shortening and the smaller the resistance, the larger the velocity of shortening.

Hopefully this help :)
7 0
3 years ago
A baseball player hits a homerun, and the ball lands in the left field seats, which is 103m away from the point at which the bal
Sati [7]

(a) The ball has a final velocity vector

\mathbf v_f=v_{x,f}\,\mathbf i+v_{y,f}\,\mathbf j

with horizontal and vertical components, respectively,

v_{x,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\cos(-38^\circ)\approx16.2\dfrac{\rm m}{\rm s}

v_{y,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\sin(-38^\circ)\approx-12.6\dfrac{\rm m}{\rm s}

The horizontal component of the ball's velocity is constant throughout its trajectory, so v_{x,i}=v_{x,f}, and the horizontal distance <em>x</em> that it covers after time <em>t</em> is

x=v_{x,i}t=v_{x,f}t

It lands 103 m away from where it's hit, so we can determine the time it it spends in the air:

103\,\mathrm m=\left(16.2\dfrac{\rm m}{\rm s}\right)t\implies t\approx6.38\,\mathrm s

The vertical component of the ball's velocity at time <em>t</em> is

v_{y,f}=v_{y,i}-gt

where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve for the vertical component of the initial velocity:

-12.6\dfrac{\rm m}{\rm s}=v_{y,i}-\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.38\,\mathrm s)\implies v_{y,i}\approx49.9\dfrac{\rm m}{\rm s}

So, the initial velocity vector is

\mathbf v_i=v_{x,i}\,\mathbf i+v_{y,i}\,\mathbf j=\left(16.2\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(49.9\dfrac{\rm m}{\rm s}\right)\,\mathbf j

which carries an initial speed of

\|\mathbf v_i\|=\sqrt{{v_{x,i}}^2+{v_{y,i}}^2}\approx\boxed{52.4\dfrac{\rm m}{\rm s}}

and direction <em>θ</em> such that

\tan\theta=\dfrac{v_{y,i}}{v_{x,i}}\implies\theta\approx\boxed{72.0^\circ}

(b) I assume you're supposed to find the height of the ball when it lands in the seats. The ball's height <em>y</em> at time <em>t</em> is

y=v_{y,i}t-\dfrac12gt^2

so that when it lands in the seats at <em>t</em> ≈ 6.38 s, it has a height of

y=\left(49.9\dfrac{\rm m}{\rm s}\right)(6.38\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.38\,\mathrm s)^2\approx\boxed{119\,\mathrm m}

6 0
3 years ago
What term best describes the regular path of a spacecraft or other object around a planetary body?
Law Incorporation [45]
B. Orbit. The planets orbit the sun, the moon orbits earth, etc.
7 0
3 years ago
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