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Liula [17]
3 years ago
12

How long does it take for a bicycle traveling 7.0 m/s to come to a stop if the

Physics
2 answers:
Olin [163]3 years ago
7 0
B: 10.5s pc.3.5sbdnshd habsbdjebs
marishachu [46]3 years ago
4 0
  • initial velocity=u=7m/s
  • Final velocity=v=0m/s
  • Acceleration=a=-3.5m/s^2

\\ \rm\Rrightarrow v=u+at

\\ \rm\Rrightarrow t=\dfrac{v-u}{a}

\\ \rm\Rrightarrow t=\dfrac{0-7}{-3.5}

\\ \rm\Rrightarrow t=\dfrac{-7}{-3.5}

\\ \rm\Rrightarrow t=2s

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Alex777 [14]

propagated disturbance is a variation

5 0
3 years ago
The summer camps had a field trip from the campus to Fragrance Hill. They traveled at an average speed of 65 km/h in the first 2
ludmilkaskok [199]

Answer:

Explanation:

They traveled this distance in 2 parts, essentially. Part 1 had an average speed for a certain number of hours, part 2 had an average speed for a certain number of hours, and those 2 parts taken together took them a distance of 364 km. In equation form, that looks like this:

km/hr part 1 + km/hr part 2 = 364 km

Now we need to find each part on the left side of that equation. Part 1 first:

We traveled 65 km/hr for 2 hours, so that took us

65\frac{km}{hr}*2hr and canceling out the hour label, we have that in part 1 we got

65(2) = 130 km. Good. Now onto the second part, where our unknown is.

We traveled 78 km/hr the second part for x hours, so that took us

78\frac{km}{hr}*xhr and canceling out the hour label, we have that in part 2 we got

78x km. Now we can fill in the main equation (the one in bold print)

130 km + 78x km = 364 km and subtracting 130 km from both sides:

78x km = 234 km and dividing by 78 km:

x = 3 hours. Part 2 took 3 hours. Part 1 took 2 hours, so the whole trip took 5 hours.

7 0
3 years ago
Why does the moon's gravity have a greater effect on the earth's ocean than the sun.
White raven [17]

Answer:

Tides on our planet are caused by the gravitational pull of the Moon and Sun. Earth's oceans "bulge out" because the Moon's gravity pulls a little harder on one side of our planet (the side closer to the Moon) than it does on the other. The Sun's gravity raises tides, too, but lunar tides are twice as big.

4 0
3 years ago
Using a 683 nm wavelength laser, you form the diffraction pattern of a 1.1 mm wide slit on a screen. You measure on the screen t
n200080 [17]

Answer:

10.2 m

Explanation:

The position of the dark fringes (destructive interference) formed on a distant screen in the interference pattern produced by diffraction from a single slit are given by the formula:

y=\frac{\lambda (m+\frac{1}{2})D}{d}

where

y is the position of the m-th minimum

m is the order of the minimum

D is the distance of the screen from the slit

d is the width of the slit

\lambda is the wavelength of the light used

In this problem we have:

\lambda=683 nm = 683\cdot 10^{-9} m is the wavelength of the light

d=1.1 mm = 0.0011 m is the width of the slit

m = 13 is the order of the minimum

y=8.57 cm = 0.0857 m is the distance of the 13th dark fringe from the central maximum

Solving for D, we find the distance of the screen from the slit:

D=\frac{yd}{\lambda(m+\frac{1}{2})}=\frac{(0.0857)(0.0011)}{(683\cdot 10^{-9})(13+\frac{1}{2})}=10.2 m

6 0
3 years ago
Calculate the heat, in kilocalories, that is absorbed if 183 g of ice at 0.0 ∘C is placed in an ice bag, melts, and warms to bod
boyakko [2]

Answer:

The total amount of heat needed will be Q_T=21.411kcal.

Explanation:

We will divide the calculation in two: First, the heat needed to melt the ice, and then the heat needed to warm the resulting liquid from 0°C to 37°C.

m=183g

l_f=80\frac{cal}{g} =334\frac{J}{g}

l_w=1\frac{cal}{g} =4.184\frac{J}{g}

<em>i) </em>The fusion heat will be:

Q_f=l_fm=14640cal=14.640kcal

<em>ii)</em> The heat needed to warm the water from T_i=0^{\circ}C to T_i=37^{\circ}C will be:

Q_w=l_wm(T_f-T_i)=6771cal=6.771kcal

So, the total amount needed will be the sum of these two results:

Q_T=Q_f+Q_w=14.640kcal+6.771kcal=21.411kcal.

8 0
3 years ago
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