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Minchanka [31]
3 years ago
11

Please help !! ASAP

Physics
1 answer:
umka2103 [35]3 years ago
5 0

Answer:

22) 5.6 Kg.m/s

23) 0.067 Kg

24) 1.16 m/s

Explanation:

Momentum is found by multiplying mass and velocity

P=mv where P is momentum, m is mass of object and v is the velocity

22

Given

Mass, m=2 Kg

Rate, v=2.8 m/s

Relationship

P=mv

Solution

P=mv=2*2.8=5.6 Kg.m/s

23

Given

Momentum, p=0.1 Kg.m/s

Rate, v=1.5 m/s

Relationship

P=mv and making m the subject then

m=\frac {P}{v}

Solution

m=\frac {0.1 Kgm.s}{1.5 m/s}\approx 0.067 Kg

24

Given

Momentum, p=4.9\times 10^{8} Kg.m/s

Mass, m=4.23\times10^{8} Kg

Relationship

P=mv and making v the subject then

v=\frac {P}{m}

Solution

m=\frac {4.9\times 10^{8}  Kgm.s}{4.23\times10^{8} }\approx 1.16 m/s

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Sophie [7]

Answer:

The fall in temperature of the liquid is 8.6 +/- 0.1 ⁰C

Explanation:

Given;

initial temperature of the liquid, t₁ = 76.3  +/-  0.4⁰C

final temperature of the liquid, t₂ = 67.7  +/-  0.3⁰C

The change in temperature of the liquid is calculated as;

Δt = t₂  -  t₁

Δt = (67.7 - 76.3)  +/-  (0.3 - 0.4)

Δt = (-8.6)  +/-  (-0.1)

Δt = 8.6 +/- 0.1 ⁰C

Therefore, the fall in temperature of the liquid is 8.6 +/- 0.1 ⁰C

4 0
3 years ago
A motorist enters a freeway at 45 km/h and accelerates uniformly to 99 km/h. From the odometer in the car, the motorist knows th
Marat540 [252]

Answer:

a)  19440 km/h²

b) 10 sec

Explanation:

v₀ = initial velocity of the car = 45 km/h

v = final velocity achieved by the car = 99 km/h

d = distance traveled by the car while accelerating = 0.2 km

a = acceleration of the car

Using the kinematics equation

v² = v₀² + 2 a d

99² = 45² + 2 a (0.2)

a = 19440 km/h²

b)

t = time required to reach the final velocity

Using the kinematics equation

v = v₀ + a t

99 = 45 + (19440) t

t = 0.00278 h

t = 0.00278 x 3600 sec

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(2.437×10⁴)(6.5411 x 10^9)/(5.37x10^6). write in scientific notation​
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Answer:

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2.968456 ×10^7

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If the magnitude of the magnetic force on a proton is F when it is moving at 18.0 ∘ with respect to the field, what is the magni
Lelu [443]

Answer:

The force when θ = 33° is 1.7625 times of the force when θ = 18°

Explanation:

The force on a moving charge through a magnetic field is given by

F = qvB sin θ

q = charge of the moving particle

v = Velocity of the moving charge

B = Magnetic field strength

θ = angle between the magnetic field and the velocity (direction of the motion) of the moving charge

Because qvB are all constant, we can call the expression K.

F = K sinθ

when θ = 18°,

F = K sin 18° = 0.309K

when θ = 33°, let the force be F₁

F₁ = K sin 33° = 0.5446K

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F₁ = 1.7625 F

Hope this Helps!!!

5 0
3 years ago
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