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Gekata [30.6K]
3 years ago
12

The eiffel tower is a steel structure whose height increases by 19.5 cm when the temperature changes from −8 to +42 °c. what is

the approximate height (in meters) at the lower temperature? m
Physics
1 answer:
Readme [11.4K]3 years ago
6 0
 The coefficient of expansion is 13 * 10^-6 m per meter length.per oK 
The temperature difference = 42 - - 8 = 50 oC 
delta T = (42 + 273) - (-8 + 273) = 50 oK 
delta L = L * 13* 10^6 m/oK 
oK = 50 oK delta L = 19.5 cm = 19.5 cm [1m / 100 cm] = 0.195m 
So we need to find the length and it is computed by:
0.195= L * 13 * 10^-6 * 50 L = 0.195 / (13*10^-6*50) L = 300 m 
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3 years ago
A 50.0 kg object is moving at 18.2 m/s when a 200 N force is applied opposite the direction of the objects motion, causing it to
Gekata [30.6K]

Answer:

t = 1.4[s]

Explanation:

To solve this problem we must use the principle of conservation of linear momentum, which tells us that momentum is conserved before and after applying a force to a body. We must remember that the impulse can be calculated by means of the following equation.

P=m*v\\or\\P=F*t

where:

P = impulse or lineal momentum [kg*m/s]

m = mass = 50 [kg]

v = velocity [m/s]

F = force = 200[N]

t = time = [s]

Now we must be clear that the final linear momentum must be equal to the original linear momentum plus the applied momentum. In this way we can deduce the following equation.

(m_{1}*v_{1})-F*t=(m_{1}*v_{2})

where:

m₁ = mass of the object = 50 [kg]

v₁ = velocity of the object before the impulse = 18.2 [m/s]

v₂ = velocity of the object after the impulse = 12.6 [m/s]

(50*18.2)-200*t=50*12.6\\910-200*t=630\\200*t=910-630\\200*t=280\\t=1.4[s]

3 0
3 years ago
A car starts from rest with an acceleration of 6
mrs_skeptik [129]

Answer:

t=16.67s

Explanation:

From the question, acceleration from 6m/s^2  \ to \ 0m/s^2 in 10 seconds.

Acceleration function can be written as:-

a_t=6-0.6t

From the acceleration equation, we can obtain the velocity equation:

v(t)=\int\limits^t_b {0} \, dv=\int\limits^t_0 {a(t)} \, dt\\v(t)=\int\limits^t_0 {(6-06t)} \, dt=6t-0.3t^2\\*dv=a(t)dt

We calculate velocity after 10sec from the above v(t) as 30m/s.

To obtain distance travelled after 10 seconds:-

S=\int\limits^{10}_0 {6t-0.3t^2} \, dt=|3t^2-0.1t^3|   \ *lims(10,0)\\ S=200m

Therefore 200m is for 10seconds, and next 200m at 30m/s

Total time=10+\frac{200}{30}=16.67s

4 0
4 years ago
What term is used to denote the change of velocity with time
Westkost [7]
The answers is acceleration.
6 0
3 years ago
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