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Vikentia [17]
3 years ago
8

Please help me I really need it

Physics
1 answer:
tensa zangetsu [6.8K]3 years ago
4 0

Answer:

A 1.5 metres per second squared B 0 C 6 0.5

b) 9m

Explanation:

a= v-a/t

a= acceleration measured in metres per second squared

v= final speed measured in metres per second

u= initial speed measured in metres per second

t= time measured in second

A

a=v-a/t

a= 3-0/2

a= 1.5 metres per second squared

Distance in a velocity time graph = area under

To calculate the distance in the first 4 seconds

You divide the graph into triangles and rectangles

Area A (triangle )= 1/2 × base × height

=1÷2 ×2×3

= 3m

Area B (rectangle)= L×B

=2×3

=6m

Area A + Area B= 3+6

= 9m

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The given velocity is 465 m/s.

Part a.
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Answer: 1674 km/h

Part b.
1674  \frac{km}{h} = (1674 \,  \frac{km}{h})*(24 \,  \frac{h}{day}  ) = 40176 \,  \frac{km}{day}
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3 years ago
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At a certain instant a particle is moving in the +x direction with momentum +8 kg m/s. During the next 0.13 seconds a constant f
jeka94

Answer:

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

Explanation:

The momentum of the particle is related to force by the following equation:

Δp = F · Δt

Where:

Δp =  change in momentum = final momentum - initial momentum

F = constant force.

Δt = time interval.

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final momentum - 8 kg m/s = -7 N · 0.13 s

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Now let´s calculate the y-component of the momentum vector after the 0.13 s. Since the particle wasn´t moving in the y-direction, the initial momentum in this direction is zero:

final momentum = 5 kg m/s² · 0.13 s

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p = (7.09 kg m/s,  0.65 kg m/s)

The magnitude of this vector is calculated as follows:

|p| = \sqrt{(7.09 kg m/s)^{2} + (0.65 kg m/s)^{2}} = 7.12 kg m/s

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

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3 years ago
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Answer:

its true

Explanation:

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Statement three i do believe

7 0
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We need the frequency of the photon, it is v = c/ λ

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