A characteristic common to waves is that they

The velocity of a 1.3 kg block sliding down a frictionless inclined plane is found to be 1.26 m/s. 1.10 s later, it has a veloci

nasty-shy [4]

Answer:

$\theta = 25.3^\circ$

Explanation:

The acceleration of the block can be found by the kinematics equations:

$v = v_0 + at\\5.88 = 1.26 + a(1.1)\\a = 4.2~m/s^2$

Since the plane is frictionless, the only force acting on the block along the motion of the block is its weight.

$F = mg\sin(\theta) = ma\\g\sin(\theta) = a\\(9.8)\sin(\theta) = 4.2\\\theta = 25.3^\circ$

7 0

3 years ago

What is a bond between a positive and a negative ion charge?

klasskru [66]

Answer:

Ionic Bond

Explanation:

5 0

3 years ago

Read 2 more answers

Why won’t anyone help me please anybody help me I really need help .

irga5000 [103]

Answer:

1➡️ this is the method of decomposition

2➡️ H2 and O2

3➡️ b

sorry if I am wrong

8 0

2 years ago

What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates

34kurt

The given question is incomplete. The complete question is as follows.

A parallel-plate capacitor has capacitance $C_{0}$ = 8.50 pF when there is air between the plates. The separation between the plates is 1.00 mm.

What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed $3.00 \times 10^{4}$ V/m?

Explanation:

It is known that relation between electric field and the voltage is as follows.

V = Ed

Now,

Q = CV

or, Q = $C \times Ed$

Therefore, substitute the values into the above formula as follows.

Q = $C \times Ed$

= $8.50 pF \times (\frac{10^{-12} F}{1 pF})(3 \times 10^{4} m/s)(1 mm)(\frac{10^{-3} m}{1 mm})$

= $2.55 \times 10^{-10} C$

Hence, we can conclude that the maximum magnitude of charge that can be placed on each given plate is $2.55 \times 10^{-10} C$ .

3 0

3 years ago

1. An express train, traveling at 36 m/s, is accidentally sidetracked onto a local train track. The express engineer spots a loc

Colt1911 [192]

Answer:

(i) 12 seconds

(ii) 216 meters from the initial position

(iii) 132 meters from the initial position

(iv) No

Explanation:

Speed of express train =36 m/s

Speed of local train =11 m/s

The initial distance between the local train and passenger train =100 m.

Due to the application of breaks, the express train slows at the rare of $3.0 m/s^2.$

So, the acceleration of the express train, $a=-3 m/s^2$ .

(i) Let t be the time the express train takes to stop.

From the equation of motion,

v=u+at

where, v: final velocity, u: initial velocity, a: constant acceleration, t: time taken to change the speed from u to v.

In this case, v=0, u=36 m/s, $a=-3 m/s^2$

So, 0=36+(-3)t

$\Rightarrow t= 36/3=12$ seconds.

(ii) To compute the distance traveled, s, till the express train stops, using

$v^2=u^2+2as$

$\Rightarrow 0^2=36^2+2(-3)s$

$\Rightarrow s=\frac{36\times36}{6}$

$\Rightarrow s=216$ meters.

(iii) The local train is moving at a speed of 11 m/s

So, in 12 seconds, the distance, d, traveled by the local train

d= 11x12=132 meters [as distance= speed x time]

(iv) Let 0 be the reference position which is the initial position of the express train.

So, at the initial time, the position of the local train is at 100m.

After 12 seconds:

The position of the express train is at 216 m [using part (ii)]

and the position of the local train is at 100+132=232m [using part (iii)].

So, the local train is still ahead of the express train, hence the trains didn't collide.

6 0

3 years ago