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Soloha48 [4]
3 years ago
9

Helppppppp pleassseeee!!!!!!

Physics
1 answer:
suter [353]3 years ago
6 0

D- Decrease use of fossil fuels.

When we burn fossil fuels we produce co2 into the air.

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Any one want to talk i'm so boreed
Sedbober [7]

Answer:

<h3>We can talk. ❤❤</h3>

Explanation:

..........

3 0
3 years ago
Write the equation expressing the relationship "y varies directly as x." use k as the constant of proportionality.
Alexeev081 [22]

The equation expressing the statement "y varies directly as x"  is y=kx.

Explanation

The statement that y varies directly as x is analogous to saying that  the ratio of y to x is constant. In other words, when x increases, y likewise increases and that when x decreases, y decreases proportionally.

Mathematically, we express the relationship that the ration of y is to x is constant is expressed as; \frac{y}{x} =k where k is the constant of proportionality.

We can then solve the relationship for y to determine the correct form of the relationship as shown below,

\frac{y}{x} =k\\\rightarrow y=kx


6 0
3 years ago
Read 2 more answers
. 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
Calculate the total resistance for a 650ohm , a 350 ohm , and a 1000 ohm resistor connected in series
Mekhanik [1.2K]

Answer:

2000 ohms

Explanation:

Resisters in series just add.

Rt = R1 + R2 + R3

R1 = 650 ohm

R2 = 350 ohm

R3 = 1000 ohm

Rt = 650 + 350 + 1000

Rt = 2000 ohms.

5 0
3 years ago
Read 2 more answers
Using Figure 2, what is the momentum of Train Car A before the collision?
Bess [88]

Answer:

Option A. 180000 Kgm/s.

Explanation:

From the question given above, the following data were obtained:

For Train Car A:

Mass of train car A = 45000 Kg

Velocity of train car A = 4 m/s

Momentum of train car A =?

For Train Car B:

Mass of train car B = 45000 Kg

Velocity of train car B = 0 m/s

Momentum is simply defined as the product of mass and velocity. Mathematically, it can be expressed as:

Momentum = mass × velocity

With the above formula, the momentum of train car A before collision can be obtained as follow:

Mass of train car A = 45000 Kg

Velocity of train car A = 4 m/s

Momentum of train car A =?

Momentum = mass × velocity

Momentum = 45000 × 4

Momentum of train car A = 180000 Kgm/s

5 0
3 years ago
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