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3 years ago

A car has positive acceleration. What information can you infer from this?

Physics

1 answer:

mario62 [17]3 years ago

6 0

Answer: B: The car is speeding up in the same direction as it is traveling.

Explanation:

I took the test and got it right

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Susan, driving north at 53 mphmph , and Shawn, driving east at 63 mphmph , are approaching an intersection. Part A What is Shawn

mafiozo [28]

Answer:

Shawn's speed relative to Susan's speed = 10 mph

Resultant velocity = 82.32 mph

Explanation:

The given data :-

i) Susan driving in north and speed of Susan is ( v₁ ) = 53 mph.

ii) Shawn driving in east and speed of Shawn is ( v₂ ) = 63 mph.

iii) The speed of both Susan and Shawn is relative to earth.

iv) The angle between Susan in north and Shawn in east is 90°.

We have to find Shawn's speed relative to Susan's speed.

v₂₁ = v₂ - v₁ = 63 - 53 = 10 mph

Resultant velocity,

$v = \sqrt{v_{2} ^{2}+ v_{1} ^{2} } =\sqrt{63^{2} +53^{2} }$

v = 82.32 mph

5 0

3 years ago

6. question is in the picture

Harlamova29_29 [7]

The answer is c you got to look for answers that make sense

8 0

4 years ago

An 80-kgkg quarterback jumps straight up in the air right before throwing a 0.43-kgkg football horizontally at 15 m/sm/s . How f

vladimir1956 [14]

Answer:

V = 0.0806 m/s

Explanation:

given data

mass quarterback = 80 kg

mass football = 0.43 kg

velocity = 15 m/s

solution

we consider here momentum conservation is in horizontal direction.

so that here no initial momentum of the quarterback

so that final momentum of the system will be 0

so we can say

M(quarterback) × V = m(football) × v (football) ........................1

put here value we get

80 × V = 0.43 × 15

V = 0.0806 m/s

5 0

3 years ago

Underground water is being pumped into a pool whose cross section is 3 m x 4 m while water is discharged through a 0.076m-diamet

Svetllana [295]

Given:

Area of pool = 3m×4m
Diameter of orifice = 0.076m
Outlet Velocity = 6.3m/s
Accumulation velocity = 1.5cm/min

Required:

Inlet flowrate

Solution:

The problem can be solved by this general formula.

Accumulation = Inlet flowrate - Outlet flowrate
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice

First, we need to convert the units of the accumulation velocity into m/s to be consistent.

Accumulation velocity = 1.5cm/min × (1min/60s)×(1m/100cm)
Accumulation velocity = 0.00025 m/s

We then calculate the area of the pool and the area of the orifice by:

Area of pool = 3 × 4 m²
Area of pool = 12m²

Area of orifice = πd²/4 = π(0.076m)²/4
Area of orifice = 0.00454m²

Since we have all we need, we plug in the values to the general equation earlier

Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice

0.00025 m/s × 12m² = Inlet flowrate - 6.3m/s × 0.00454m²

Transposing terms,

Inlet flowrate = 0.316 m³/s

6 0

3 years ago

If you decrease the distance between successive crests of a wave, this changes

AlexFokin [52]

<span>The _______ is the the distance between two crests or two troughs on a transverse wave. It is also the distance between compressions or the distance between rarefactions on a longitudinal wave.</span>

3 0

3 years ago