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OLga [1]
3 years ago
12

In which situation are both the speed and velocity of the car changing?

Physics
2 answers:
lubasha [3.4K]3 years ago
6 0
Answer is A because the speed and velocity would change. Think of it as GTA, your going 30+ miles per hour and you take a left turn, the speed and velocity would change in an instant.. 
Hope this helped.
NemiM [27]3 years ago
6 0
The answer to this question is:
<span>In which situation are both the speed and velocity of the car changing?
A) The car turns left at 5
m
s
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
If I made a mistake please feel free to tell me ^-^
</span>
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What is the main reason why people use machines
Butoxors [25]

Answer:

Machines are faster and more efficent

Explanation:

5 0
2 years ago
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. Consider the equation =0+0+02/2+03/6+04/24+5/120, where s is a length and t is a time. What are the dimensions and SI units of
Olegator [25]

Answer:

See Explanation

Explanation:

Given

s=s_0+v_0t+\frac{a_0t^2}{2}+ \frac{j_0t^3}{6}+\frac{S_0t^4}{24}+\frac{ct^5}{120}

Solving (a): Units and dimension of s_0

From the question, we understand that:

s \to L --- length

t \to T --- time

Remove the other terms of the equation, we have:

s=s_0

Rewrite as:

s_0=s

This implies that s_0 has the same unit and dimension as s

Hence:

s_0 \to L --- dimension

s_o \to Length (meters, kilometers, etc.)

Solving (b): Units and dimension of v_0

Remove the other terms of the equation, we have:

s=v_0t

Rewrite as:

v_0t = s

Make v_0 the subject

v_0 = \frac{s}{t}

Replace s and t with their units

v_0 = \frac{L}{T}

v_0 = LT^{-1}

Hence:

v_0 \to LT^{-1} --- dimension

v_0 \to m/s --- unit

Solving (c): Units and dimension of a_0

Remove the other terms of the equation, we have:

s=\frac{a_0t^2}{2}

Rewrite as:

\frac{a_0t^2}{2} = s_0

Make a_0 the subject

a_0 = \frac{2s_0}{t^2}

Replace s and t with their units [ignore all constants]

a_0 = \frac{L}{T^2}\\

a_0 = LT^{-2

Hence:

a_0 = LT^{-2 --- dimension

a_0 \to m/s^2 --- acceleration

Solving (d): Units and dimension of j_0

Remove the other terms of the equation, we have:

s=\frac{j_0t^3}{6}

Rewrite as:

\frac{j_0t^3}{6} = s

Make j_0 the subject

j_0 = \frac{6s}{t^3}

Replace s and t with their units [Ignore all constants]

j_0 = \frac{L}{T^3}

j_0 = LT^{-3}

Hence:

j_0 = LT^{-3} --- dimension

j_0 \to m/s^3 --- unit

Solving (e): Units and dimension of s_0

Remove the other terms of the equation, we have:

s=\frac{S_0t^4}{24}

Rewrite as:

\frac{S_0t^4}{24} = s

Make S_0 the subject

S_0 = \frac{24s}{t^4}

Replace s and t with their units [ignore all constants]

S_0 = \frac{L}{T^4}

S_0 = LT^{-4

Hence:

S_0 = LT^{-4 --- dimension

S_0 \to m/s^4 --- unit

Solving (e): Units and dimension of c

Ignore other terms of the equation, we have:

s=\frac{ct^5}{120}

Rewrite as:

\frac{ct^5}{120} = s

Make c the subject

c = \frac{120s}{t^5}

Replace s and t with their units [Ignore all constants]

c = \frac{L}{T^5}

c = LT^{-5}

Hence:

c \to LT^{-5} --- dimension

c \to m/s^5 --- units

4 0
3 years ago
1. A diver dives off of a raft - what happens to the diver? The raft? How does this relate to Newton's Third Law?  Action Force:
Natali [406]
The Action Force of this scenario is the pushing force of the Diver. The Reaction Force is the raft pushing back on the diver. 

The Third Law of Motion states that "For every action, there is an equal and opposite reaction." Now when the diver dives off the raft, the raft is also pushing the same amount of force as the diver did as he dives off. The diver will then move forward and the raft on the other hand will move backwards.

The movement of the raft shows the opposite force. It will move backwards depending on how strong the diver will push off on the raft. And the amount of force he pushes on it, the raft will exert the same force so the stronger the force of the diver, the farther he will go because the raft will push him in that same direction as it goes backwards. 
7 0
3 years ago
Please help quickly!
Naddik [55]

Answer:

○ D

Explanation:

A resistor is a passive electrical component that uses a circuit element with electric resistance. Resistors are used in electronics for reducing current flow, changing signal rates, dividing voltages, biasing active components, and terminating columns, among other applications.

I am joyous to assist you anytime.

7 0
3 years ago
How much physical activity do adults need?
serg [7]
30? Physical activity
8 0
2 years ago
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