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2 years ago

Positively charged particles consisting of two protons and two neutrons emitted by radioactive materials are

Physics

2 answers:

dangina [55]2 years ago

7 0

Your answer is alpha particles

antiseptic1488 [7]2 years ago

4 0

Answer:

alpha particles

Explanation:

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a car accelerates from rest at 3.6m/s^2. how much time does it need to attain a speed of 7 m/s? awnser in units of s.

Ivahew [28]

Initial velocity: 0
final velocity: 7 m/s
a = 3.6
t = ?
x = ?

(7-0)/3.6 = t
t = 1.94 s

6 0

3 years ago

Read 2 more answers

Four equal masses m are so small they can be treated as points, and they are equallyspaced along a long, stiff mass less wire. T

gavmur [86]

The moment of inertia of a point mass about an arbitrary point is given by:

I = mr²

I is the moment of inertia

m is the mass

r is the distance between the arbitrary point and the point mass

The center of mass of the system is located halfway between the 2 inner masses, therefore two masses lie ℓ/2 away from the center and the outer two masses lie 3ℓ/2 away from the center.

The total moment of inertia of the system is the sum of the moments of each mass, i.e.

I = ∑mr²

The moment of inertia of each of the two inner masses is

I = m(ℓ/2)² = mℓ²/4

The moment of inertia of each of the two outer masses is

I = m(3ℓ/2)² = 9mℓ²/4

The total moment of inertia of the system is

I = 2[mℓ²/4]+2[9mℓ²/4]

I = mℓ²/2+9mℓ²/2

I = 10mℓ²/2

I = 5mℓ²

4 0

2 years ago

If you push a crate with a force of 100 N and it slides at a constant speed and in the same direction, how much friction is acti

Nadusha1986 [10]

Answer: 100 N

Explanation: Taking into account the second Newton Law the total force applied to any system is equal to the mass *acceleration.In this case the crate moves at constant speed so the accelaration is zero. In order to satisfy this fact, the friction force must be equal the applied force of 100 N .

8 0

3 years ago

Middle C, also known as do on the fixed do-solfge scale, is a note used when playing or singing music. The frequency of middle C

Ahat [919]

Answer:

The speed of sound for middle C is 345.01 m/s.

Explanation:

Given that,

The frequency of middle C is 261.63 Hz

The wavelength of middle C note used when playing or singing music, $\lambda=131.87\ cm=1.3187\ m$

We need to find the speed of sound for middle C. We know that the relation between speed, frequency and wavelength is given by the formula as follows :

$v=f\lambda\\\\v=261.63\ Hz\times 1.3187\ m\\\\v=345.01\ m/s$

So, the speed of sound for middle C is 345.01 m/s. Hence, this is the required solution.

8 0

3 years ago

Let the masses of blocks A and B be 4.50 kg and 2.00 kg , respectively, the moment of inertia of the wheel about its axis be 0.4

Free_Kalibri [48]

Answer:

Accelerations of both the sides is 0.6125 $m/s^{2}$ , A moves downwards whereas B moves upwards.

$\alpha$ =6.125 rad/ $s^{2}$

Tension on side A = 4.5 × g= 44.1 m/ $s^{2}$

Tension on side B= 2.0 × g= 19.6 m/ $s^{2}$

Explanation:

As both, the blocks A and B are attached due to the constraint they can only possess a single acceleration a.

Observe the figure attached, let the tension with Block A be $T_{2}$ and the tension attached with Block B be $T_{1}$ .

Tensions will be only be due to the weight of the blocks as no other force is present.

$T_{2}$ = 4.5 × g= 44.1 m/ $s^{2}$

$T_{1}$ = 2.0 × g= 19.6 m/ $s^{2}$

Now, lets make a torque equation about the center of the wheel and find the alpha

$T_{2}$ ×R- $T_{1}$ ×R= MI( Moment of Inertia of Wheel)× Alpha

where, R= Radius of the wheel=0.100m and

Alpha( $\alpha$ )= Angular acceleration of the wheel

MI of the wheel= 0.400 kg/ $m^{2}$

$(44.1-19.6)R=0.400\alpha$

$\alpha = \frac{24.5 * 0.100}{0.400}$

$\alpha$ =6.125 rad/ $s^{2}$

Acceleration = R × $\alpha$

= 0.1 * 6.125

=0.6125 $m/s^{2}$

Accelerations of both the sides is 0.6125 $m/s^{2}$ , A moves downwards whereas B moves upwards.

7 0

3 years ago