physics mcqs Watch

HUSSAI55
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https://pastpapers.papacambridge.com..._s18_qp_13.pdf
can someone explain q7,14,16,28,
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Meowstic
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7: The slope must be steep enough so the vertical component of M's gravitational acceleration wins out
14: pressure = force per unit area
16: what are you always told to ignore in these questions that you haven't been told to ignore here
28: half the constant acceleration in the same time, think suvat
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HUSSAI55
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for 14 why cant it be pressure exerted by the gas?
and i still dont get 28 thanks tho.
(Original post by HUSSAI55)
https://pastpapers.papacambridge.com..._s18_qp_13.pdf
can someone explain q7,14,16,28,
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HUSSAI55
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https://pastpapers.papacambridge.com..._w17_qp_11.pdf
can someone explain 14 ,33,36,37 ?
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Teenie2
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(14) Work done = force x distance. As Meowstic says, pressure = force/area. The work done is against the atmospheric pressure.
(33) R = ρL/A
(36) If the total resistance in the circuit reduces then what happens to the current?
(37) How many volts are there across the 160ohm resistor? This should give you the current and from that you can work out R.
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HUSSAI55
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oh this is another exam paper i cant do the above questions from this paper thanks for ur help tho
oct 2017 p11 cie
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Eimmanuel
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(Original post by HUSSAI55)
for 14 why cant it be pressure exerted by the gas?
and i still dont get 28 thanks tho.
For Q14, you need to realize that the gas pressure inside is changing NOT constant.
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Eimmanuel
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(Original post by HUSSAI55)
for 14 why cant it be pressure exerted by the gas?
and i still dont get 28 thanks tho.
Q28

If the particle of mass m and charge +q enters a uniform electric field with speed v, the magnitude of the acceleration of the particle is qE/m.

The second particle of mass 2m, charge +q and speed v enters the electric field along the same would have a magnitude of the acceleration as  \dfrac{qE}{2m} . This is half the magnitude of the acceleration of the first particle, so the deflection would be half of d as the time taken for the second particle to travel through the region of the electric region is same.

The time taken to travel through the region of the electric field is determined by the velocity perpendicular to the electric field and this is the same in both cases.
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