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Lemur [1.5K]
2 years ago
5

The car's initial speed was 32m / s and it takes 99 seconds for the car to come to a full stop after the driver applies the brak

es. What is the magnitude of the car's acceleration?
Physics
1 answer:
Juli2301 [7.4K]2 years ago
8 0

Answer:

0.323232....m/s2

Explanation:

acceleration=V/t

32m/s/99s=0.3232m/s2

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ss7ja [257]

B. They are rearranged.

The First Law of Thermodynamics states that matter and energy can not be created or destroyed.

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3 years ago
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An 8.0 cm object is 40.0 cm from a concave mirror that has a focal length of 10.0 cm. Its image is 16.0 cm in front of the mirro
svet-max [94.6K]
 We can rearrange the mirror equation before plugging our values in. 
1/p = 1/f - 1/q. 
1/p = 1/10cm - 1/40cm
1/p = 4/40cm - 1/40cm = 3/40cm
40cm=3p  <-- cross multiplication
13.33cm = p

Now that we have the value of p, we can plug it into the magnification equation.

M=-16/13.33=1.2
1.2=h'/8cm
9.6=h'

So the height of the image produced by the mirror is 9.6cm.
6 0
2 years ago
Can anyone solve these for my by using unit vectors? Can you also please show your work
Oxana [17]

4. The Coyote has an initial position vector of \vec r_0=(15.5\,\mathrm m)\,\vec\jmath.

4a. The Coyote has an initial velocity vector of \vec v_0=\left(3.5\,\frac{\mathrm m}{\mathrm s}\right)\,\vec\imath. His position at time t is given by the vector

\vec r=\vec r_0+\vec v_0t+\dfrac12\vec at^2

where \vec a is the Coyote's acceleration vector at time t. He experiences acceleration only in the downward direction because of gravity, and in particular \vec a=-g\,\vec\jmath where g=9.80\,\frac{\mathrm m}{\mathrm s^2}. Splitting up the position vector into components, we have \vec r=r_x\,\vec\imath+r_y\,\vec\jmath with

r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)t

r_y=15.5\,\mathrm m-\dfrac g2t^2

The Coyote hits the ground when r_y=0:

15.5\,\mathrm m-\dfrac g2t^2=0\implies t=1.8\,\mathrm s

4b. Here we evaluate r_x at the time found in (4a).

r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)(1.8\,\mathrm s)=6.3\,\mathrm m

5. The shell has initial position vector \vec r_0=(1.52\,\mathrm m)\,\vec\jmath, and we're told that after some time the bullet (now separated from the shell) has a position of \vec r=(3500\,\mathrm m)\,\vec\imath.

5a. The vertical component of the shell's position vector is

r_y=1.52\,\mathrm m-\dfrac g2t^2

We find the shell hits the ground at

1.52\,\mathrm m-\dfrac g2t^2=0\implies t=0.56\,\mathrm s

5b. The horizontal component of the bullet's position vector is

r_x=v_0t

where v_0 is the muzzle velocity of the bullet. It traveled 3500 m in the time it took the shell to fall to the ground, so we can solve for v_0:

3500\,\mathrm m=v_0(0.56\,\mathrm s)\implies v_0=6300\,\dfrac{\mathrm m}{\mathrm s}

5 0
3 years ago
Someone help please by providing work and answers please :)
Nastasia [14]
First we gotta use an equation of motion:

d = ut + \frac{1}{2} a {t}^{2}

Our vertical distance d= 100 m, initial vertical speed u = 0 m/s (because velocity is fully horizontal), and vertical acceleration a = 9.8 m/s2 because of gravity. Let's plug it all in!

100 = 0 + \frac{1}{2} (9.8) {t}^{2}

Now we just need to solve for t:

{t}^{2} = \frac{2(100)}{9.8} \\ \\ t = \sqrt{\frac{2(100)}{9.8}}

Hit the calculators, and you'll get 4.5 seconds!
5 0
3 years ago
What's is meant by relative motion?​
saveliy_v [14]

Relative motion means a motion relative to a reference point. We can also say, relative motion means motion referred or observed from a reference point.

For example, Alex is in a train and Ace is at the station, when the train starts moving, for Ace it is moving at a speed of 10 m/s, but for Alex it is moving at 0 m/s, or we can say that it is at rest for Alex, but in motion for Ace. This is relative motion.

4 0
3 years ago
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