Suppose you increase your walking speed from 5 m/s to 11 m/s in a period of 1 s. What is your acceleration?

On a hypothetical scale X The ice point is 40° and steam point is 120°.

arlik [135]

Answer:

The reading of Y is -10°.

Explanation:

For scale X, the ice point is 40° and steam point is 120°.

Difference between the two extremes for scales X = 120 - 40 = 80

For scale X, the ice point and steam points are -30° and 130° respectively.

Difference between the two extremes for scales X = 130 - (-30) = 160

Comparing both scales:

One unit of scale X = x

One unit of scale Y = y

Scale X has 80 divisions while scale Y has 160

80x = 160y

x = 2y

50° in scale X = 10x + ice point in X scale

10 divisions in Y scale = 20y

Reading of Y scale = ice point of Y + 20y

= -30° + 20°

= -10°

7 0

3 years ago

Wegener developed he hypothesis that Earth's landmasses had once been fused together, and hen slowly broke apart in a process ca

ivanzaharov [21]

Solution: C. Pangea

The hypothesis of Continental drift suggests that in past, there was only one landmass on the Earth -Pangaea which drifted apart due to movement of plate tectonics causing earthquake. This hypothesis is supported by many evidences which includes presence of similar fossils on different landmasses, common land features such as widespread presence of glacial sediments etc.

4 0

3 years ago

An ion's position vector is initially r with arrow = 7.0i hat â 7.0j + 1.0k, and 5.0 s later it is r with arrow = 7.0i hat + 7.0

harkovskaia [24]

<h2>Answer:</h2>

Shown in the explanation

<h2>Explanation:</h2>

Position vector of a particle at a given instant is given by:

$\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$

On the other hand, the average velocity is the change in the particle’s position vector divided by time interval:

$\vec{v}=\frac{\Delta \vec{r}}{\Delta t}=\frac{\vec{r_{2}}-\vec{r_{1}} }{t_{2}-t_{1}} \\ \\ \\ Where: \\ \\ \\ \vec{r_{1}}: Initial \ position \\ \\ \vec{r_{2}}: Final \ position \\ \\ t_{1}:Initial \ time \\ \\ t_{2}:Final \ time$

So we have:

$\vec{r_{1}}=7\hat{i}+7\hat{j}+1\hat{k} \\ \\ \vec{r_{2}}=7\hat{i}+7\hat{j}+8\hat{k} \\ \\ \\ Then: \\ \\ \Delta \vec{r} = \vec{r_{2}}-\vec{r_{1}}=7\hat{i}+7\hat{j}+8\hat{k}-(7\hat{i}+7\hat{j}+1\hat{k}) \\ \\ \Delta \vec{r}=7\hat{k} \\ \\ \\ We \ also \ know: \\ \\ \Delta t=5s$

Finally, the average velocity is:

$\vec{v}=\frac{\Delta \vec{r}}{\Delta t} \\ \\ \\ \vec{v}=\frac{7\hat{k}}{5} \\ \\ \boxed{\vec{v}=\frac{7}{5}\hat{k} \ units/s}$

3 0

4 years ago

(A) What is the maximum tension possible in a 1.00-millimeter-diameter nylon tennis racket string?

attashe74 [19]

Complete Question

(A) What is the maximum tension possible in a 1.00- millimeter-diameter nylon tennis racket string?

(B) If you want tighter strings, what do you do to prevent breakage: use thinner or thicker strings? Why? What causes strings to break when they are hit by the ball?

The tensile strength of the nylon string is $600*10^{6} \ N/m^2$