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deff fn [24]
3 years ago
10

An athlete runs 40 meters north and 30 meters south

Physics
1 answer:
Serggg [28]3 years ago
7 0

Answer:

only ten meters north from starting point

Explanation:

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An aluminum cylinder with a radius of 2.7 cm and a height of 67 cm is used as one leg of a workbench. The workbench pushes down
soldier1979 [14.2K]

Answer:

1.9\times 10^{-4}

1.2\times 10^{-4}\ m

Explanation:

r = Radius = 2.7 cm

F = Force = 3.2\times 10^4\ N

A = Area = \pi r^2

\sigma = Stress = \frac{F}{A}

E = Young's modulus = 7\times 10^{10}\ Pa

\epsilon = Strain

L_0 = Original length = 67 cm

\Delta L = Change in length

Young's modulus is given by

E=\frac{\sigma}{\epsilon}\\\Rightarrow \epsilon=\frac{\sigma}{E}\\\Rightarrow \epsilon=\frac{\frac{3.2\times 10^4}{\pi 0.027^2}}{7\times 10^{10}}\\\Rightarrow \epsilon=0.0001996=1.9\times 10^{-4}

Strain is 1.9\times 10^{-4}

Strain is given by

\epsilon=\frac{\Delta L}{L_0}\\\Rightarrow \Delta L=\epsilon\times L_0\\\Rightarrow \Delta L=1.9\times 10^{-4}\times 0.67\\\Rightarrow \Delta L=0.0001273\\\Rightarrow \Delta L=1.2\times 10^{-4}\ m

The cylinder height decreases by 1.2\times 10^{-4}\ m

3 0
3 years ago
A woman is walking at 4 m/s. She is accelerating at a rate of 1 m/s2. To find out what her velocity is after 3 seconds, what els
liq [111]

Answer:

distance

Explanation:

8 0
2 years ago
Read 2 more answers
Two boats leave the same port at the same time, with boat A traveling north at 15 knots (nautical miles per hour) and boat B tra
Mrrafil [7]

Answer:

The chance in distance is 25 knots

Explanation:

The distance between the two particles is given by:

s^2 = (x_A - x_B)^2+(y_A - y_B)^2  (1)

Since A is traveling north and B is traveling east we can say that their displacement vector are perpendicular and therefore (1) transformed as:

s^2 = x_B^2+y_A^2 (2)

Taking the differential with respect to time:

\displaystyle{2s\frac{ds}{dt}= 2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt}}  (3)

where \displaystyle{\frac{dx_B}{dt}}=v_B and \displaystyle{\frac{dx_A}{dt}}=v_A are the respective given velocities of the boats. To find s and x_B we make use of the given position for A, y_A=30, the Pythagoras theorem and the relation between distance and velocity for a movement with constant velocity.

\displaystyle{y_A = v_A\cdot t\rightarrow t = \frac{y_A}{v_A}=\frac{30}{15}=2 h

with this time, we know can now calculate the distance at which B is:

\displaystyle{x_B = v_B\cdot t= 20 \cdot 2 = 40\ nmi

and applying Pythagoras:

\displaystyle{s = \sqrt{x_B^2+y_A^2}=\sqrt{30^2 + 40^2}=\sqrt{2500}=50}

Now substituting all the values in (3) and solving for  \displaystyle{\frac{ds}{dt} } we get:

\displaystyle{\frac{ds}{dt} = \frac{1}{2s}(2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt})}\\\displaystyle{\frac{ds}{dt} = 25 \ knots}

4 0
3 years ago
NEWTONS SECOND LAW LAB REPORT
hram777 [196]

Answer:

Newton's second law states that the acceleration of an object is directly related to the net force and inversely related to its mass. Acceleration of an object depends on two things, force and mass.

Explanation:

here this may help.

8 0
3 years ago
Two boxes are at rest on a smooth, horizontal surface. The boxes are in contact with one another. If box 1 is pushed with a forc
Maslowich

Answer:

mass of box 1 = 2.20 kg

mass of box 2 = 5.93 kg

Explanation:

Let the mass of box 1 and box 2 is respectively

m_1 and m_2

so we will have

Force applied on box 1 then acceleration

a = \frac{F}{m_1 + m_2}

1.50 = \frac{12.2}{m_1 + m_2}

m_1 + m_2 = 8.13

Now we know that contact force between them in above case is given as

F_n = m_2a

8.90 = m_2(1.5)

m_2 = 5.93 kg

now we have

m_1 = 2.20 kg

8 0
3 years ago
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