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ahrayia [7]
4 years ago
14

If you were to drop a rock from a tall building, assuming that it had not yet hit the ground, and neglecting air resistance. Wha

t is its vertical displacement (in m) after 6 s?
Physics
1 answer:
Snowcat [4.5K]4 years ago
3 0

Answer:

176.4 m

Explanation:

U = 0, t = 6s, g = 9.8 m/s^2

Use second equation of motion

H = ut + 1/2 gt^2

H = 0 + 0.5 × 9.8 × 6 × 6

H = 176.4 m

It is the displacement from the point of dropping of object.

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Unlike mechanical waves such as sound or earthquake waves, EM waves can travel through empty space. True or false
Alex777 [14]

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its true

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5 0
3 years ago
How much net force is needed to accelerate a 15 kg mass at 2.8 m/s^2
Lera25 [3.4K]
Refer to the 2th Law of Newton

F = m. a

F = 15 x 2.8 = 42 N
4 0
3 years ago
Express the length of a marathon run of<br> 26 mi 385 yd in meters.<br> Answer in units of m.
hjlf

Answer:

1 mi = 5280 ft  * 12 in/ft = 63360 in

A convenient conversion factor (to remember) is 1 m = 39.37 in

63360 in / (39.37 in / m) = 1609.3 m

26 mi + 285 m = 26 * 1609.3 + 385 = 42,228 m

6 0
3 years ago
Consider a simple tension member that carries an axial load of P=22.44N. Find the total elongation in the member due to the load
rodikova [14]

Answer:

The total elongation for the tension member is of 0.25mm

Explanation:

Assuming that material is under a linear deformation then the relation between the stress and the specific elongation is given as:

\sigma=E*\epsilon (1)

Where E is the modulus of elasticity, σ the stress and ε the specific deformation. Also, the total longitudinal elongation can be expressed as:

\delta L=L*\epsilon (2)

Here L is the member extension and δL the change total longitudinal elongation.  

Now if the stress is found then the deformation can be calculated by solving the stress-deformation equation (1). The stress applied sigama is computed dividing the axial load P by the cross-sectional area A:

\sigma=P/A  

\sigma=22.44N / 1290 mm^2  

\sigma=0.0174 N/mm^2  

Solving for epsilon and replacing the calculated value for the stress and the value for the modulus of elasticity:

=\sigma=E*\epsilon

\epsilon=\sigma/E

\epsilon=0.0174 \frac{N}{mm^2}/\ 204 \frac{N}{mm^2}

\epsilon=8.53*10^-{5}

Finally introducing the specific deformation and the longitudinal extension in the equation of total elongation (2):

\delta L=3048 mm * 8.53*10^{-5}  

\delta L= 0.25 mm

4 0
3 years ago
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