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Korvikt [17]
2 years ago
12

Kon'nichiwa~please help me with this question!!​

Physics
1 answer:
andrezito [222]2 years ago
4 0

Let's check the relationship

\\ \rm\hookrightarrow g=\dfrac{GM}{r^2}

\\ \rm\hookrightarrow g\propto G

So

  • Raindrops will fall faster . .
  • Also walking on ground would become more difficult as g increases.

Option C is wrong by now .Let's check D once

\\ \rm\hookrightarrow T\propto \dfrac{1}{\sqrt{g}}

  • So time period of simple pendulum would decrease.

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In 1965, a group of students wore black arm bands to school in protest of American policies in Vietnam. Administrators banned th
Ede4ka [16]

Answer:

protected under students first amendment rights

Explanation:

did the studyisland :)

4 0
2 years ago
Please help me this is important!
Mariulka [41]
145 Grams!
It asks for the “Total Mass” basically asking to add, If you add 20 to 125, you get 145! Correct me if im wrong
4 0
3 years ago
You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo; there is negligible friction between
Anni [7]

Answer:

0.074m/s

Explanation:

We need the formula for conservation of momentum in a collision, this equation is given by,

m_1u_1+m_2u_2 = m_1v_1+m_2v_2

Where,

m_1 = mass of ball

m_2 = mass of the person

u_1 = Velocity of ball before collision

u_2 = Velocity of the person before collision

v_1 = velocity of ball afer collision

v_2= velocity of the person after collision

We know that after the collision, as the person as the ball have both the same velocity, then,

v_1 = v_2

m_1u_1 + m_2u_2 = (m_1+m_2)v_2

Re-arrenge to find v_2,

v_2 = \frac{m_1u_1+m_2u_2}{m_1+m_2}

Our values are,

m_1= 0.425kg

u_1= 12m/s

m_2= 68.5kg

u_2= 0m/s

Substituting,

v_2 = \frac{(0.425)(12)+(68.5)(0)}{0.425+68.5}

v_2 = 0.074m/s

<em />

<em>The speed of the person would be 0.074m/s after the collision between him/her and the ball</em>

7 0
3 years ago
An electric furnace runs 13 hours a day to heat a house during January (31 days). The heating element has a resistance of 7.2 an
liq [111]

Answer:

cost of running the furnace during January is $5619.62

Explanation:

given data

runs a day = 13 hours

January days = 31 days

resistance = 7.2 ohm

current = 16.7 A

cost of electricity = $0.10/kWh

to find out

cost of running the furnace during January

solution

first we get her power consumed by furnace that is

Power consumed = \frac{I^2}{R}  ........1

put here value we get

Power consumed = \frac{16.7^2}{7.2}

Power consumed = 38.7347 W

and

Power consumed by furnace in one hour is

Power consumed by furnace in one hour is = Power consumed × 3600

Power consumed by furnace in one hour is = 38.7347 × 3600  

Power consumed by furnace in one hour is 139.445kWh

and

Power consumed by furnace in the month of January is

Power consumed by furnace in the month of January = 139.445kWh × 13 hours × 31 days

Power consumed by furnace in the month of January = 56196.335 kWh

so

cost of running the furnace during January is = $0.10/kWh × 56196.335 kWh

cost of running the furnace during January is $5619.62

4 0
3 years ago
Read 2 more answers
How would I find the average distance for this?
Maslowich

Answer:

Explanation:add them then divide them by 100 I think

3 0
2 years ago
Read 2 more answers
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