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777dan777 [17]
3 years ago
15

A car has a kinetic energy of 4.32 x 10^5 J when traveling at a speed of 23 m/s.

Physics
1 answer:
Kisachek [45]3 years ago
4 0

Answer:

I think the most interesting 3rd century 3g durga was 5f5

Explanation:

I 50degree and I have not done a good morning mam sushant since I am not pregnant and have no

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The rigid beam is supported by the three suspender bars. bars ab and ef are made of aluminum and bar cd is made of steel. if eac
faltersainse [42]

Answer:

Pmax = 67.5 KN

Explanation:

We need to calculate the maximum allowable value of P for both aluminum and steel bars.

<u>FOR STEEL BARS</u>:

Since,

(σallow)st = (Pmax)st/A

where,

(σallow)st = maximum allowable stress of steel bar = 200 MPa = 2 x 10⁸ Pa

A = Cross-sectional area of steel bar = 450 mm² = 0.45 x 10⁻³ m²

(Pmax)st = Maximum allowable force for steel bar = ?

Therefore,

2 x 10⁸ Pa = (Pmax)st/0.45 x 10⁻³ m²

(Pmax)st = (2 x 10⁸ Pa)(0.45 x 10⁻³ m²)

(Pmax)st = 9 x 10⁴ N = 90 KN

<u>FOR Aluminum BARS</u>:

Since,

(σallow)al = (Pmax)al/A

where,

(σallow)al = maximum allowable stress of Al bar = 150 MPa = 1.5 x 10⁸ Pa

A = Cross-sectional area of Aluminum bar = 450 mm² = 0.45 x 10⁻³ m²

(Pmax)al = Maximum allowable force for Aluminum bar = ?

Therefore,

1.5 x 10⁸ Pa = (Pmax)al/0.45 x 10⁻³ m²

(Pmax)al = (1.5 x 10⁸ Pa)(0.45 x 10⁻³ m²)

(Pmax)al = 6.75 x 10⁴ N = 67.5 KN

Since,

(Pmax)al < (Pmax)st

Therefore,

The maximum allowable force will be:

Pmax = (Pmax)al

<u>Pmax = 67.5 KN</u>

3 0
4 years ago
When performing an.experiment similar to Millikan's oil drop, a student measured the following load magnitudes: 3.26x10 ^-19 C 5
Scilla [17]

Answer:

1.6 x 10⁻¹⁹ C

Explanation:

Let us arrange the charges in the ascending order and round them off as follows :-

1.53 x 10⁻¹⁹ C   → 1.6x 10⁻¹⁹ C

3.26 x 10⁻¹⁹C   → 3.2 x 10⁻¹⁹ C

4.66 x 10⁻¹⁹C   → 4.8 x 10⁻¹⁹ C

5.09 x 10⁻¹⁹C   → 4.8 x 10⁻¹⁹ C

6.39 x 10⁻¹⁹C   → 6.4 x 10⁻¹⁹ C

The rounding off has been made to facilitate easy calculation to come to a conclusion and to accommodate error in measurement.

Here we observe that

2 nd charge is almost twice the first charge

3 rd and 4 th charges are almost 3 times the first charge

5 th charge is almost 4 times the first charge.

This result implies that 2 nd to 5 th charges are made by combination of the first charge ie if we take e as first  charge , 2nd to 5 th charges can be  written as 2e,  3e ,3e and 4e. Hence e is the minimum charge existing in nature and on electron this minimum charge of  1.6 x 10⁻¹⁹ C  exists.

3 0
3 years ago
Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.55 times a second. A tack is stuck in the tire a
earnstyle [38]

Answer:

Tangential speed = 5.72 m/s

Centripetal acceleration = 91.6\text{ m/s}{}^2

Explanation:

The tangential speed, V, is given by

v=\omega r

where \omega is the angular speed and is given by 2\pi f (f is the angular frequency or frequency of rotation)

Thus,

v=2\pi f r = 2\times3.14\times2.55\times0.357 = 5.72\text{ m/s}

The centripetal acceleration,a, is given by

a=\dfrac{v^2}{r}

a=\dfrac{5.72^2}{0.357} = 91.6\text{ m/s}{}^2

7 0
4 years ago
Read 2 more answers
I need help ! if anyone answers this its worth 40 points please help !!
Sliva [168]
Double displacement...I think
3 0
3 years ago
The angular velocity of a process control motor is (13−12t2) rad/s, where t is in seconds. Part A At what time does the motor re
mihalych1998 [28]

Answer:

Explanation:

Given

\omega =13-\frac{1}{2}\cdot t^2

Motor reverse its direction when \omega =0

13-0.5t^2=0

26=t^2

t=\sqrt{26}=5.099\approx 5.1 s

(b)

\frac{\mathrm{d} \theta }{\mathrm{d} t}=\omega

\int d\theta =\int_{0}^{5.1}\omega dt

\int d\theta =\int_{0}^{t}(13-.05t^2)dt

\theta =(13t-0.1667\times t^3)_0^{5.1}

\theta =44.192^{\circ}

4 0
3 years ago
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