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stepan [7]
3 years ago
13

Larry sees a group of people weeping, with frowns on their faces and their eyes turned down. Larry their expressions to understa

nd that they are feeling sadness.
Select one:
a. misinterprets
b. decodes
C. encodes X
d. recodes​
Physics
1 answer:
Salsk061 [2.6K]3 years ago
8 0

Answer:

c

Explanation:

no need explanation u can trust me

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Particle 1 and particle 2 have masses of m1 = 2.2 × 10-8 kg and m2 = 4.8 × 10-8 kg, but they carry the same charge q. The two pa
Lorico [155]

Answer:r_2=11.81 cm

Explanation:

Given

m_1=2.2\times 10^{-8} kg

m_2=4.8\times 10^{-8} kg

same charge on both masses

potential Energy due to Magnetic Field =Kinetic Energy of Particle

qV=\frac{mv^2}{2}

v=\sqrt{\frac{2qV}{m}}

and we know

Force due to magnetic field will Provide centripetal Force

qvB=\frac{mv^2}{r}

B=\frac{\sqrt{\frac{2Vm}{q}}}{r}

and B is equal for both particles

thus \frac{m}{r^2}=constant

\frac{m_1}{r_1^2}=\frac{m_2}{r_2^2}

r_2^2=\frac{4.8}{2.2}\times 8^2

r_2=11.81 cm

4 0
3 years ago
A ray diagram shows an object placed between 2F and F of a convex lens.The image produced is
nekit [7.7K]
The image produced is magnified and real.
4 0
3 years ago
Read 2 more answers
A car accelerates from rest to 19 m/s in 6 seconds. What is the acceleration of the car?
choli [55]

Answer:

the car is moving so that how it gos so fast

Explanation:

7 0
3 years ago
A diver leaves the end of a 4.0 m high diving board and strikes the water 1.3s later, 3.0m beyond the end of the board. Consider
shutvik [7]

Answer:

4.0 m/s

Explanation:

The motion of the diver is the motion of a projectile: so we need to find the horizontal and the vertical component of the initial velocity.

Let's consider the horizontal motion first. This motion occurs with constant speed, so the distance covered in a time t is

d=v_x t

where here we have

d = 3.0 m is the horizontal distance covered

vx is the horizontal velocity

t = 1.3 s is the duration of the fall

Solving for vx,

v_x = \frac{d}{t}=\frac{3.0 m}{1.3 s}=2.3 m/s

Now let's consider the vertical motion: this is an accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. The vertical position at time t is given by

y(t) = h + v_y t - \frac{1}{2}gt^2

where

h = 4.0 m is the initial height

vy is the initial vertical velocity

We know that at t = 1.3 s, the vertical position is zero: y = 0. Substituting these numbers, we can find vy

0=h+v_y t - \frac{1}{2}gt^2\\v_y = \frac{0.5gt^2-h}{t}=\frac{0.5(9.8 m/s^2)(1.3 s)^2-4.0 m}{1.3 s}=3.3 m/s

So now we can find the magnitude of the initial velocity:

v=\sqrt{v_x^2+v_y^2}=\sqrt{(2.3 m/s)^2+(3.3 m/s)^2}=4.0 m/s

4 0
3 years ago
Assume that the position vector of A is r=i+j+k . Determine the moment about the origin O if the force F=(1)i+(0)j+(5)k . The mo
ddd [48]

Answer:

M₀ = 5i - 4j - k

Explanation:

Using the cross product method, the moment vector(M₀) of a force (F) is about a given point is equal to cross product of the vector A from the point (r) to anywhere on the line of action of the force itself. i.e

M₀ = r x F

From the question,

r = i + j + k

F = 1i + 0j +  5k

Therefore,

M₀ = (i + j + k) x (1i + 0j +  5k)

M₀ = \left[\begin{array}{ccc}i&j&k\\1&1&1\\1&0&5\end{array}\right]

M₀ = i(5 - 0) -j(5 - 1) + k(0 - 1)

M₀ = i(5) - j(4) + k(-1)

M₀ = 5i - 4j - k

Therefore, the moment about the origin O of the force F is

M₀ = 5i - 4j - k

3 0
3 years ago
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