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zysi [14]
3 years ago
5

V

Physics
1 answer:
Darina [25.2K]3 years ago
4 0
Resolving vectors question. See diagram enclosed.

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What medical technique uses sound wave pulses and their reflections to map structures and organs inside the body?
garri49 [273]

The correct answer is “C” ultrasound. Hope this helps!

7 0
4 years ago
Read 2 more answers
A small block with a mass of 0.0600 kg is attached to a cord passing through a hole in a frictionless, horizontal surface (Figur
pogonyaev

Answer with Explanation:

We are given that mass of block=0.0600 kg

Initial speed of block=0.63 m/s

Distance of block  from the hole when the block is revolved=0.47 m

Final speed=3.29 m/s

Distance of block  from the hole when the block is revolved=9\times 10^{-2}m

a.We have to find the tension in the cord in the original situation when the block has speed =v_0=0.63 m/s

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

Because tension is equal to centripetal force

Substitute the values

T=\frac{0.06\times (0.63)^2}{0.47}=0.05 N

b.v=3.29 m/s

T=\frac{mv^2}{r}=\frac{0.06\times (3.29)^2}{0.09}=7.2 N

c.Work don=Final K.E-Initial K.E

W=\frac{1}{2}m(v^2-v^2_0)

W=\frac{1}{2}(0.06)((3.29)^2-(0.63)^2)

W=0.31 J

4 0
3 years ago
Read 2 more answers
Calculate the work against gravity required to build the right circular cone of height 4 m and base of radius 1.2 m out of a lig
Nana76 [90]

Answer:

Work done = 35467.278 J

Explanation:

Given:

Height of the cone = 4m

radius (r) of the cone = 1.2m

Density of the cone = 600kg/m³

Acceleration due to gravity, g = 9.8 m/s²

Now,

The total mass of the cone (m) = Density of the cone × volume of the cone

Volume of the cone = \frac{1}{3}\pi r^2 h

thus,

volume of the cone = \frac{1}{3}\pi 1.2^2\times 4 = 6.03 m³

therefore, the mass of the cone = 600 Kg/m³ × 6.03 m³ = 3619.11 kg

The center of mass for the cone lies at the \frac{1}{4}times the total height

thus,

center of mass lies at,  h' = \frac{1}{4}\times4=1m

Now, the work gone (W) against gravity is given as:

W = mgh'

W = 3619.11kg × 9.8 m/s² × 1 = 35467.278 J

4 0
3 years ago
A system absorbs 194 kj of heat and the surroundings do 120 kj of work on the system. internal eneergy change
notsponge [240]
We can solve the problem by using the first law of thermodynamics, which states that:
\Delta U = Q-W
where
\Delta U is the change in internal energy of the system
Q is the heat absorbed by the system
W is the work done by the system

In our problem, the heat absorbed by the system is Q=+194 kJ, while the work done is W=-120 kJ, where the negative sign means the work is done by the surroundings on the system. Therefore, the variation of internal energy is
\Delta U= Q-W=+194 kJ - (-120 kJ)=+314 kJ
6 0
3 years ago
The visible surface of the Sun, which is also the deepest layer we can see with our eyes, is in which region of the Sun?
Nina [5.8K]
It is the photosphere
5 0
3 years ago
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