Momentum before the hit:
p = mv = 0.01 * 300 + 1 * 0

Momentum after the hit:
p = 0.01 * 150 + 1 * v

Momentum is conserved:
0.01 * 300 = 0.01 * 150 + v
3 = 1.5 + v
v = 1.5

The velocity of the block after the collision is 1.5 m/s.

7 0

4 years ago

A 15 kg block is on a ramp which is inclined at 20o above the horizontal. It is connected by a string to a 19 kg mass which hang

12345 [234]

Answer:

The magnitude of the acceleration of the 19 kg block is 1.414 m/s²

Explanation:

From Newton's second law of motion;

$F_{Net} = ma$

where;

m is the mass of the objects involved, kg

a is the acceleration of the object, m/s²

different forces on the block and string

⇒force due to 15 kg block

=mgcosθ = 15×9.8×cos20 = 15×9.8×0.9396

= 138.12 N

⇒Tensional Force on 19 kg mass:

T = mg = 19×9.8 = 186.2 N

$F_{Net} = T-mg = a(m_1+m_2)$

186.2 - 138.12 = a(15+19)

48.08 = a(34)

a = 48.08/34

a = 1.414 m/s²

Therefore, the magnitude of the acceleration of the 19 kg block is 1.414 m/s²

7 0

4 years ago

The vertical displacement of the wave is measured from the

algol13

The vertical displacement of the wave is measured from
the equilibrium to the crest and is called the amplitude.

6 0

4 years ago

Read 2 more answers

The MRI (Magnetic Resonance Imaging) body scanners used in hospitals operate at a frequency of 400 MHz (M = 106 ). Calculate the

horrorfan [7]

Answer:

Energy, $E=2.65\times 10^{-25}\ J$

Explanation:

It is given that,

The MRI (Magnetic Resonance Imaging) body scanners used in hospitals operate at a frequency of 400 MHz,

$\nu=400\ MHz=400\times 10^6\ Hz=4\times 10^8\ Hz$

We need to find the energy for a photon having this frequency. The energy of a photon is given by :

$E=h\nu$

$E=6.63\times 10^{-34}\ J-s\times 4\times 10^8\ Hz$

$E=2.65\times 10^{-25}\ J$

So, the energy of the photon is $2.65\times 10^{-25}\ J$ . Hence, this is the required solution.

6 0

3 years ago