Which of the following was NOT one of Ghana's chief trading exports?

A body moving in the positive x direction passes the origin at time t = 0. Between t=0 and t=1 second, the body has a constant s

lord [1]

The important thing in this type of questions is to keep track of what you are doing at every point.

From t=0 to t=1s, there is uniform motion at v=24 m/s.

Therefore, it will move 24m in that second of time.

Afterwards, an acceleration comes in, so from t=1s to 11s there will be uniformly accelerated motion with acceleration -6m/s^2.

To account for this, you need to use Suvat's equation:

x(t)=x0+v0*t+1/2*a*t^2.

You should know what to plug in for each of the symbols in this equation:

x0 [initial position at t=1s] = 24m

v0 [initial velocity at t=1s] = 24 m/s

a [acceleration, switched on at t=1s] = -6 m/s^2

t [time from the start of the acceleration until the end, i.e. from t=1s to t=11s] = 10s

Plugging in those numbers in the equation will give you the position at t=11s.

4 0

3 years ago

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Give an example of a positive slope

Mekhanik [1.2K]

Answer:

uu yt? gcc5feesserhujnggbgxf

Explanation:

c

6 0

3 years ago

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All valid equations in physics have consistent units. Are all equations that have consistent units valid?

sweet [91]

Answer:

c. No. An equation may have consistent units but still be numerically invaid.

Explanation:

For an equation to be corrected, it should have consistent units and also be numerically correct.

Most equation are of the form;

(Actual quantity) = (dimensionless constant) × (dimensionally correct quantity)

From the above, without the dimensionless constant the equation would be numerically wrong.

For example; Kinetic energy equation.

KE = 0.5(mv^2)

Without the dimensionless constant '0.5' the equation would be dimensionally correct but numerically wrong.

8 0

3 years ago

In Speed Study Number 1, we looked at two cars traveling the same distance at different speeds on city streets. Car "A" traveled

Ganezh [65]

Answer:

230.4 s

Explanation:

The speed of car A is

$v_A = 35 mi/h$

and the distance travelled is

$d = 10 mi$

so the time taken for car A is

$t_A = \frac{d}{v_A}=\frac{10 mi}{35 mi/h}=0.286 h$

The speed of car B is

$v_B = 45 mi/h$

and the distance travelled is

$d = 10 mi$

so the time taken for car B is

$t_B = \frac{d}{v_B}=\frac{10 mi}{45 mi/h}=0.222 h$

So the difference in time is

$\Delta t = t_A - t_B = 0.286 h -0.222 h=0.064 h$

Which corresponds to

$\Delta t = 0.064 h \cdot 3600 s/h = 230.4 s$

so car B arrived 230.4 s before car A.

5 0

2 years ago

Can anyone tell me the ans of this last question quick please

tia_tia [17]

Explanation:

Speed: distance/time

Average speed: total distance/total time

Total distance: 400

Total time: 60

Average spead: 400/60= 6.67m/s

8 0

2 years ago