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3 years ago

Which characteristic does an object with a constant acceleration always have?

Physics

2 answers:

9966 [12]3 years ago

8 0

Explanation:

By definition, the word acceleration is equal to the rate of change of velocity. Mathematically, it is given by :

$a=\dfrac{dv}{dt}$

$dv=a.dt$

$v=\int\limits^t_0 {a.dt}$

Since, it is given that acceleration is constant

$v=at+v_o$

v₀ is the constant of integration and it corresponds to initial velocity

From above equation, it is clear that when acceleration is constant the speed varies linearly. Hence, when an object move with constant acceleration, it always changes its velocity.

lapo4ka [179]3 years ago

5 0

It always has changing velocity

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As frequency increases in an electromagnetic wave, its photon energy decreases. A. True B. False

Diano4ka-milaya [45]

Answer:

False

Explanation:

3 0

4 years ago

A uniform line charge of density λ lies on the x axis between x = 0 and x = L. Its total charge is 7 nC. The electric field at x

DedPeter [7]

Answer:

The electric field at x = 3L is 166.67 N/C

Solution:

As per the question:

The uniform line charge density on the x-axis for x, 0< x< L is $\lambda$

Total charge, Q = 7 nC = $7\times 10^{- 9} C$

At x = 2L,

Electric field, $\vec{E_{2L}} = 500N/C$

Coulomb constant, K = $8.99\times 10^{9} N.m^{2}/C^{2}$

Now, we know that:

$\vec{E} = K\frac{Q}{x^{2}}$

Also the line charge density:

$\lambda = \frac{Q}{L}$

Thus

Q = $\lambda L$

Now, for small element:

$d\vec{E} = K\frac{dq}{x^{2}}$

$d\vec{E} = K\frac{\lambda }{x^{2}}dx$

Integrating both the sides from x = L to x = 2L

$\int_{0}^{E}d\vec{E_{2L}} = K\lambda \int_{L}^{2L}\frac{1}{x^{2}}dx$

$\vec{E_{2L}} = K\lambda[\frac{- 1}{x}]_{L}^{2L}] = K\frac{Q}{L}[frac{1}{2L}]$

$\vec{E_{2L}} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{2L}] = \frac{63}{L^{2}}$

Similarly,

For the field in between the range 2L< x < 3L:

$\int_{0}^{E}d\vec{E} = K\lambda \int_{2L}^{3L}\frac{1}{x^{2}}dx$

$\vec{E} = K\lambda[\frac{- 1}{x}]_{2L}^{3L}] = K\frac{Q}{L}[frac{1}{6L}]$

$\vec{E} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{6L}] = \frac{63}{6L^{2}}$

Now,

If at x = 2L,

$\vec{E_{2L}} = 500 N/C$

Then at x = 3L:

$\frac{\vec{E_{2L}}}{3} = \frac{500}{3} = 166.67 N/C$

4 0

4 years ago

The day VFR visibility and cloud clearance requirements to operate over the town of Cooperstown, after departing and climbing ou

pshichka [43]

Answer:

The correct option is;

C. 1 mile clear of clouds

Explanation:

Given that the indicated airspace location is at or below 700 feet AGL therefore, it is taken as being in the region of a class G airspace which covers the airspace regions from the base up to and equal to 1,200 feet beneath the class E airspace and the requirement for VFR flight for class G are 1 mile and clear of clouds.

4 0

3 years ago

What is the net torque on the square plate, with sides 0.2 m, from each of the three forces? F1=18 N, F2=26 N, and F3=14 N. Use

marshall27 [118]

Answer:

The net torque on the square plate is 2.72 N-m.

Explanation:

Given that,

Side = 0.2 m

Force $F_{1}=18\ N$