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3 years ago

G A dragster starts from rest and accelerates at 35 m/s2 m / s 2 . How fast is it going after t t

Physics

1 answer:

bija089 [108]3 years ago

4 0

Complete question:

A dragster starts from rest and accelerates at 35 m/s2 m / s 2 . How fast is it going after 7s.

Answer:

The final velocity of the dragster is 245 m/s.

Explanation:

Given;

acceleration of the dragster, a = 35 m/s²

initial velocity of the dragster, u = 0

time of motion, t = 7 s

the final velocity of the dragster after 7s is given by;

v = u + at

where;

v is the final velocity

u is the initial velocity

v = 0 + (35 x 7)

v = 245 m/s

Therefore, the final velocity of the dragster is 245 m/s.

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Explanation:

They will repel, meaning that they are made of an electrical conductor.

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3 years ago

A spring is hung from the ceiling. A 2.0-kg mass suspended hung from the spring extends it by 6.0 cm. A downward external force

Stolb23 [73]

The work done by force on a spring hung from the ceiling will be 1.67 J

Any two things with mass are drawn together by the gravitational pull. We refer to the gravitational force as attractive because it consistently seeks to draw masses together rather than pushing them apart.

Given that a spring is hung from the ceiling with a 2.0-kg mass suspended hung from the spring extends it by 6.0 cm and a downward external force applied to the mass extends the spring an additional 10 cm.

We need to find the work done by the force

Given mass is of 2 kg

So let,

F = 2 kg

x = 0.1 m

Stiffness of spring = k = F/x

k = 20/0.006 = 333 n/m

Now the formula to find the work done by force will be as follow:

Workdone = W = 0.5kx²

W = 0.5 x 333 x 0.1²

W = 1.67 J

Hence the work done by force on a spring hung from the ceiling will be 1.67 J

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brainly.com/question/12970081

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2 years ago

Consider the sections of two circuits illustrated above. Select True or False for all statements. After connecting c and d to a

Basile [38]

Answer:

Follows are the solution to the given question:

Explanation:

In this question the missing file of the circuit is not be which is defined in attached file please find it.

In Option 1, this statement is true because the current is on $R_3 \ and \ R_4$ , that is the same.

In option 2, this statement is false because $Rcd=R_3+R_4$ therefore it implies that Rcd is always larger then $R_3$ .

In option 3, this statement is true because the voltage of $R_1 \ and \ R_2$ is always equal.

In option 4, this statement is true because $Rab= \frac{1}{(\frac{1}{R1}+\frac{1}{R2})}=\frac{R1R2}{(R1+R2)}. \frac{R2}{(R1+R2)}$ is always smaller then 1 therefore, $R_1 \times (\frac{R2}{(R1+R2)})$ is always equal to R1.