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topjm [15]
3 years ago
9

North America experienced all of the following during the last glacial period EXCEPT alpine glaciers covered the Rocky and Casca

de Mountains. sea level was at least 100 m lower than it is today. coastlines were extended. the Hudson Bay was covered with alpine glaciers.
Physics
1 answer:
Juli2301 [7.4K]3 years ago
5 0

Answer:

the Hudson Bay was covered with alpine glaciers

Explanation:

During the last glacial period, large portions of North America were covered with ice. The majority of the ice was from the ice sheets that were covering Canada and the northern part of the United States, and the alpine glaciers on the mountain ranges. Hudson Bay was all frozen at this point of time. It was not covered with alpine glaciers though, instead it was covered with the ice of the extended ice sheets, with the ice cover reaching up to 2 km in thickness.

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Vector A has y-component Ay= +15.0 m . A makes an angle of 32.0 counterclockwise from the +y-axis. What is the x component of A?
Gemiola [76]

Answer:

x-component=-9.3 m

Magnitude of A=17.7m

Explanation:

We are given that

A_y=+15 m

\theta=32^{\circ}

We have to find the x-component of A and magnitude of A.

According to question

A_y=\mid A\mid cos\theta

Substitute the values then we get

15=\mid A\mid cos32

\mid A\mid =\frac{15}{cos32}=\frac{15}{0.848}

\mid A\mid=17.7m

tan\theta=\frac{perpendicular\;side}{Base}

tan32=\frac{A_x}{A_y}=\frac{A_x}{15}

0.62\times 15=A_x

A_x=9.3

The value of x-component of A is negative because the vector A lie  in second quadrant.

Hence, the x- component of A=-9.3 m

6 0
3 years ago
Read 2 more answers
A 0.106-A current is charging a capacitor that has square plates 6.00 cm on each side. The plate separation is 4.00 mm. (a) Find
FrozenT [24]

Answer:

The time rate of change of flux is 1.34 \times 10^{10} \frac{V}{s}

Explanation:

Given :

Current I = 0.106 A

Area of plate A = 36 \times 10^{-4} m^{2}

Plate separation d = 4 \times 10^{-3} m

(A)

First find the capacitance of capacitor,

   C = \frac{\epsilon _{o} A }{d}

Where \epsilon _{o} = 8.85 \times 10^{-12}

   C = \frac{8.85 \times 10^{-12 } \times 36 \times 10^{-4}  }{4 \times 10^{-3} }

   C = 7.9 \times 10^{-12} F

But   C = \frac{Q}{V}

Where Q = It

  C = \frac{It}{V}

  V = \frac{It}{C}

Now differentiate above equation wrt. time,

  \frac{dV}{dt} = \frac{I}{C}

       = \frac{0.106}{7.9 \times 10^{-12} }

       = 1.34 \times 10^{10} \frac{V}{s}

Therefore, the time rate of change of flux is 1.34 \times 10^{10} \frac{V}{s}

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3 years ago
Read the scenario below and answer the question that follows. Randall is hiring cooks for his restaurant. The first applicant is
Talja [164]
Randall has unconscious assumption that attractive people are more competent
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3 years ago
A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the
Softa [21]

Answer:

Q=3.9825\times 10^{-9} C

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S=\frac{360}{10000}=0.036 m^2

\Delta d=0.8 cm=0.008 m

\Delta V=100 V

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

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C=\frac{\epsilon_0 S}{d}

Capacitance of capacitor after moving plates

C_1=\frac{\epsilon_0 S}{(d+\Delta d)}

V=\frac{Q}{C}

Potential difference between plates after moving

V=\frac{Q}{C_1}

V+\Delta V=\frac{Q}{C_1}

\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}

\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100

\frac{Q\Delta d}{\epsilon_0 S}=100

\epsilon_0=8.85\times 10^{-12}

Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}

Q=3.9825\times 10^{-9} C

Hence, the charge on positive plate of capacitor=Q=3.9825\times 10^{-9} C

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3 years ago
Need help finding the average speed.
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Explanation:

To find the average of these numbers, we just have to add the three numbers together and divide by 3.

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