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3 years ago

14

you walk forward 10 meters, backwards 5 meters, and then 10 meters forwards in 15 seconds. What is your velocity FORWARDS in m/s

1 answer:

nata0808 [166]3 years ago

8 0

Your velocity forward in would be 1.66 m/s. to solve you would use the velocity equation.

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Two rockets are flying in the same direction and are side by side at the instant their retrorockets fire. Rocket A has an initia

Sedaia [141]

Answer:

$a_2\ =\ -33.65\ m/s^2$

Explanation:

Given,

For the first rocket,

Initial velocity of the first rocket A = $u_1\ =\ 4600\ m/s.$

Acceleration of the first rocket = $a_1\ =\ -18\ m/s^2$

For the second rocket,

Initial velocity of the second rocket B = $u_2\ =\ 8200 m/s.$
Displacement of both the rockets A and B = s = 0 m

Fro the first rocket,

Let 't' be the time taken by the first rocket A for whole the displacement

$\therefore s\ =\ u_1t\ +\ \dfrac{1}{2}a_1t^2\\\Rightarrow 0\ =\ 4600t\ -\ 0.5\times 18t^2\\\Rightarrow t\ =\ \dfrac{4600}{0.5\times 18}\\\Rightarrow t\ =\ 511.11 sec$

Let $a_2$ be the acceleration of the second rocket B for the same time interval

from the kinematics,

$\therefore s\ =\ ut\ +\ \dfrac{1}{2}at^2\\\Rightarrow s\ =\ u_2t\ +\ \dfrac{1}{2}a_2t^2\\\Rightarrow a_2\ =\ \dfrac{2s\ -\ 2u_2t}{t^2}\\\Rightarrow a_2\ =\ \dfrac{0\ -\ 2u_2t}{t^2}\\$

$\Rightarrow a_2\ =\ -\dfrac{2u_2}{t}\\\Rightarrow a_2\ =\ -\dfrac{2\times 8600}{511.11}\\\Rightarrow a_2\ =\ -33.65\ m/s^2$

Hence the acceleration of the second rocket B is -33.65\ m/s^2.

6 0

3 years ago

What is the current through an electric heater with a resistance of 38  when the potential difference is 240 V?

Zinaida [17]

$\bf{ \underline{Given:- }}$

$\sf• \: Resistance \: (R) \: = 38 \: Ω$

$• \sf \: Potential \: difference \: (V) \: = 240 \: v$

$\bf{ \underline{To \: Find:- }}$

$• \sf \: The \: current \: through \: an \: electric \: heater.$

<h2> $\bf{ \underline{ Solution :-}}$ </h2>

$\bf \red{•\: The \: formula \: of \: Current \: (I) = \frac{V}{R}}$

$\sf \rightarrow I = \frac{240}{38}$

$\sf \rightarrow I = 6.32$

$\sf \pink{Answer :- \: The \: current \: through \: an \: electric \: heater \: is \: 6.32 \: A.}$

6 0

3 years ago

Two forces act on a 6.00-kg object. One of the forces is 10.0 N. If the object accelerates at 2.00 m/s2

liubo4ka [24]

Given :

Two forces act on a 6.00-kg object. One of the forces is 10.0 N.

Acceleration of object 2 m/s².

To Find :

The greatest possible magnitude of the other force.\

Solution :

Let, other force is f.

So, net force, F = 10 + f.

Now, acceleration is given by :

$a=\dfrac{F}{mass}\\\\a= \dfrac{10+f}{6}\\\\\dfrac{10+f}{6}=2\\\\f = 12 - 10\\\\f = 2 \ N$

Therefore, the greatest possible magnitude of the other force is 2 N.

Hence, this is the required solution.

7 0

3 years ago

If the energy input of an engine is 6,000 J and its efficiency is 33%, calculate its energy output. Eout = ______ J 200 2000 300

AnnyKZ [126]

The answer to the question is 2,000 J. Hope this helps!

6 0

4 years ago

Read 2 more answers

An object is moving with an initial velocity of 19 m/s.It is then subject to a constant acceleration of 2.5 m/s for 15s.How far

bija089 [108]

Answer:

566.3 m

Explanation:

The distance travelled by the object can be found by using the SUVAT equation:

$d=ut+\frac{1}{2}at^2$

where

u is the initial velocity

t is the time

a is the acceleration

For the object in this problem:

u = 19 m/s

a = 2.5 m/s^2

Substituting t = 15 s, we find the distance travelled:

$d=(19)(15)+\frac{1}{2}(2.5)(15)^2=566.3 m$

7 0

4 years ago