The resultant of two vectors is maximum, when angle between them is:

A wave with a period of 0.008 second has a frequency of

coldgirl [10]

The frequency of a wave is the reciprocal of its period.

A period of 0.008 sec means a frequency of

1 / 0.008 sec = 125 per sec . (125 Hz)

8 0

3 years ago

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In a game of angry birds you launch a bird with an angle of 53 degrees to horizontal. Unfortunatly, its not a good shot and the

Alisiya [41]

Answer:

The maximum height covered is 3.25 m.

The horizontal distance covered is 9.81 m.

The total time in the air is 1.63 seconds.

Explanation:

The launch speed, $u_0= 10 m/s$ .

Angle of launch with the horizontal, $\theta = 53 ^{\circ}$

So, the vertical component of the initial velocity,

$u_0\sin\theta=10 \sin 53 ^{\circ}\cdots(i)$ .

The horizontal component of the initial velocity,

$u_0\cos\theta=10 \cos 53 ^{\circ}$

Let, t be the time of flight, to the horizontal distance covered

$D=10 \cos (53 ^{\circ})t\cdots(ii)$ .

Not, applying the equation of motion in the vertical direction.

$s= ut +\frac 1 2 at^2$

Where s is the displacement in time t, u is the initial velocity and a is the acceleration.

In this case, $u =10 \sin 53 ^{\circ}$ (from equation (i), s=0 (as the final height is same as the launch height) and $a = -9.81 m/s^2$ (negative sign is due to the downward direction).

$\Rightarrow 0 = 10 (\sin 53 ^{\circ})t-\frac 1 2 (9.81)t^2$

$\Rightarrow t= \frac {2\times 10 (\sin 53 ^{\circ})}{9.81}=1.63$ seconds.

So, the total time in the air is 1.63 seconds.

From equation (i),

Total horizontal distance covered is

$D=10 \cos (53 ^{\circ})\times 1.63 = 9.81 m$ .

Now, for the maximum height, H, applying the equation of motion as

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

Here, v is the final velocity and v=0 (at the maximum height), and h=H.

So, $0^2=(10 \sin 53 ^{\circ})^2-2(9.81)H$

$\Rightarrow H = \frac {(10 \sin 53 ^{\circ})^2}{2\times 9.81}$

$\Rightarrow H = 3.25 m$ .

Hence, the maximum height covered is 3.25 m.

8 0

2 years ago

The linear momentum of a car of mass 1000 kg moving with a speed of 10 m/s is-----kg.m/s. HELP ME

Assoli18 [71]

Answer:

4

Explanation:

The kilogram-meter per second (kg · m/s or kg · m · s -1 ) is the standard unit of momentum . Reduced to base units in the International System of Units ( SI ), a kilogram-meter per second is the equivalent of a newton-second (N · s), which is the SI unit of impulse .

5 0

3 years ago

What causes the coriolis effect

Marta_Voda [28]

C is the correct answer

7 0

2 years ago

A ball is thrown vertically upwards from the edge of the cliff and hits the ground at the base of the cliff with a speed of 30 m

olya-2409 [2.1K]

To solve this problem we will apply the linear motion kinematic equations. From the definition of the final velocity, as the sum between the initial velocity and the product between the acceleration (gravity) by time, we will find the final velocity. From the second law of kinematics, we will find the vertical position traveled.

$v = v_0 -gt$