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Anna71 [15]
3 years ago
6

One fine summer day a group of students was jumping from a railroad bridge into the river below. They stepped off the bridge wit

h an acceleration (due to gravity) of 9.8 m/s2 and they hit the water 1.5 s later. How high (in meters) was the bridge? Only enter the number, not the units.
Physics
1 answer:
Mrrafil [7]3 years ago
3 0

Answer:

11 m

Explanation:

The following data were obtained from the question:

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) = 1.5 s

Height (h) =..?

The height of the bridge can be obtained by using the following formula:

h =½gt²

h = ½ × 9.8 × 1.5²

= 4.9 × 2.25

= 11.025 ≈ 11 m

Thus, the height of the bridge is approximately 11 m

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7 0
3 years ago
A strip of copper 150 µm thick and 4.50 mm wide is placed in a uniform magnetic field of magnitude B = 0.74 T, that is perpendic
Sophie [7]

Answer:

V = 6.55*10^{-6} v

Explanation:

The number density can be determined by using below formula:

n = \frac{Bi}{Vle}

where,

B  is uniform magnetic field 0.74

i is current 18 A

V is hall potential difference

l is thickness 150 MICRO METER

e is electron charge 1.6 *10^{-19} C

therefore V can be determined as

V = \frac{iB}{nle}

V = \frac{18*0.74}{8.47*10^{28}*150*10^{-6}*1.6 *10^{-19}}

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8 0
3 years ago
What would happen to the seasons if the Earth’s tilt was only 10 degrees and why? What about 50 degrees and why?
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If it was only 10 degrees the seasonal changes would lessen and vise versa for 50. meaning the seasonal changes would become more extreme

Explanation:

8 0
3 years ago
Observations of the Crab Nebula taken over several decades show that gas blobs that are now 100 arcseconds from the center of th
Oksi-84 [34.3K]

Answer:

Estimation: year 1110.

Explanation:

We need to know how much time it takes to move 100 arcseconds if it moves at 0.11 arcsecond per year. Similarly to any velocity equation v=\frac{d}{t}, where in our case the distances are angular, we will obtain the time by doing:

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6 0
3 years ago
A cake is removed from a 350◦F oven and placed on a cooling rack in a 70◦F room. After 30 minutes the cake is 200◦F. When will i
galben [10]

Answer:

350 F to 100 F it take approx 87.33 min  

Explanation:

given data

oven = 350◦F

cooling rack = 70◦F

time = 30 min

cake = 200◦F

solution

we apply here Newtons law of cooling  

\frac{dT}{dt} = -k(T-Ta)

\frac{dy}{dt} = \frac{d}{dt} (T(t) -Ta)

= \frac{dT}{dt} -\frac{dTa}{dt} =\frac{dT}{dt} = -k(T-Ta)

-ky \frac{dy}{dt} = -ky

T(t) -Ta = (To -Ta) e^{-kt} T(t) = Ta+ (To -Ta)  e^{-kt}

put her value for time 30 min and T(t) = 200◦F and To =350◦F  and Ta = 70◦F

so here

200 = 70 + ( 350 - 70 ) e^{-k30}

k = 0.025575

so here for  T(t) = 100F

100 = 70 + ( 350 - 70 ) e^{-0.025575*t}

time = 87.33 min

so here 350 F to 100 F it take approx 87.33 min  

5 0
4 years ago
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