Light can travel from the sun to ________ in less than ten minutes.

The 5-kg block A has an initial speed of 5 m/s as it slides down the smooth ramp, after which it collides with the stationary bl

Akimi4 [234]

Answer:

The coupled velocity of both the blocks is 1.92 m/s.

Explanation:

Given that,

Mass of block A, $m_1=5\ kg$

Initial speed of block A, $u_1=5\ m/s$

Mass of block B, $m_2=8\ kg$

Initial speed of block B, $u_2=0$

It is mentioned that if the two blocks couple together after collision. We need to find the common velocity immediately after collision. We know that due to coupling, it becomes the case of inelastic collision. Using the conservation of linear momentum. Let V is the coupled velocity of both the blocks. So,

$m_1u_1+m_2u_2=(m_1+m_2)V\\\\V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{5\times 5+0}{(5+8)}\\\\V=1.92\ m/s$

So, the coupled velocity of both the blocks is 1.92 m/s. Hence, this is the required solution.

8 0

2 years ago

Help me, i don't know the answer

8_murik_8 [283]

Explanation:

1)

A) Bb BB

B) 50%

2)

A) 50%

B) b. b

B. Bb. Bb

b. bb. bb

4 0

2 years ago

SOH-CAH-TOA is used to solve for the ________

Misha Larkins [42]

Answer:

c. initial (x and y)

Explanation:

When a projectile is launched at a velocity with a launch angle, to solve it, we must first resolve the initial velocity into the x and y components. To do this will mean we have to treat it like a triangle due to the launch angle and the direction of the projectile.

Therefore, we will have to make use of trigonometric ratios which is also known by the mnemonic "SOH CAH TOA"

Thus, this method resolves the initial x and y velocities.

3 0

2 years ago

PLEASE PROVIDE AN EXPLANATION. THANKS!!!

ziro4ka [17]

Answer:

(a) A = 0.0800 m, λ = 20.9 m, f = 11.9 Hz

(b) 250 m/s

(c) 1250 N

(d) Positive x-direction

(e) 6.00 m/s

(f) 0.0365 m

Explanation:

(a) The standard form of the wave is:

y = A cos ((2πf) t ± (2π/λ) x)

where A is the amplitude, f is the frequency, and λ is the wavelength.

If the x term has a positive coefficient, the wave moves to the left.

If the x term has a negative coefficient, the wave moves to the right.

Therefore:

A = 0.0800 m

2π/λ = 0.300 m⁻¹

λ = 20.9 m

2πf = 75.0 rad/s

f = 11.9 Hz

(b) Velocity is wavelength times frequency.

v = λf

v = (20.9 m) (11.9 Hz)

v = 250 m/s

(c) The tension is:

T = v²ρ

where ρ is the mass per unit length.

T = (250 m/s)² (0.0200 kg/m)

T = 1250 N

(d) The x term has a negative coefficient, so the wave moves to the right (positive x-direction).

(e) The maximum transverse speed is Aω.

(0.0800 m) (75.0 rad/s)

6.00 m/s

(f) Plug in the values and find y.

y = (0.0800 m) cos((75.0 rad/s) (2.00 s) − (0.300 m⁻¹) (1.00 m))

y = 0.0365 m

8 0

2 years ago

Read 2 more answers

2. Suppose that a parallel-plate capacitor has circular plates with radius R = 30 mm and a plate separation of d = 5.0 mm. Suppo

seropon [69]

Answer:

The maximum value of the induced magnetic field is $2.901\times10^{-13}\ T$ .

Explanation:

Given that,

Radius of plate = 30 mm

Separation = 5.0 mm

Frequency = 60 Hz

Suppose the maximum potential difference is 100 V and r= 130 mm.

We need to calculate the angular frequency

Using formula of angular frequency

$\omega=2\pi f$

Put the value into the formula

$\omega=2\times\pi\times60$

$\omega=376.9\ rad/s$

When r>R, the magnetic field is inversely proportional to the r.

We need to calculate the maximum value of the induced magnetic field that occurs at r = R

Using formula of magnetic filed

$B_{max}=\dfrac{\mu_{0}\epsilon_{0}R^2\timesV_{max}\times\omega}{2rd}$

Where, R = radius of plate

d = plate separation

V = voltage

Put the value into the formula

$B_{max}=\dfrac{4\pi\times10^{-7}\times8.85\times10^{-12}\times(30\times10^{-3})^2\times100\times376.9}{2\times130\times10^{-3}\times5.0\times10^{-3}}$

$B_{max}=2.901\times10^{-13}\ T$

Hence, The maximum value of the induced magnetic field is $2.901\times10^{-13}\ T$ .

7 0

3 years ago