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2 years ago

True or false, question in picture,

Physics

2 answers:

lakkis [162]2 years ago

8 0

Answer:

its false

Explanation:

Arlecino [84]2 years ago

3 0

Answer:

false

Explanation:

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How to find average speed in physics

Triss [41]

Answer: you divide total distance by time. To get the time, divide total distance by speed. To get distance, multiply speed times the amount of time.

Explanation:

I hope this helps

7 0

2 years ago

Define anterior and posterior in correlation to the body. (ANATOMY)

tatyana61 [14]

Answer:

it should be right it's from go.ogle hm!!!

Explanation:

Anterior or ventral - front (example, the kneecap is located on the anterior side of the leg). Posterior or dorsal - back (example, the shoulder blades are located on the posterior side of the body). Medial - toward the midline of the body (example, the middle toe is located at the medial side of the foot).

8 0

2 years ago

Read 2 more answers

A substance with a definite shape and volume is a<br> phase of matter

lisabon 2012 [21]

Answer:

Explanation:

solid

This type of matter is know as solid. Also, it is known that solid substances have definite shape and volume. Thus, we can conclude that a substance with a definite shape and volume is a solid phase of matter.

3 0

3 years ago

Read 2 more answers

Two friction disks A and B are brought into contact when the angular velocity of disk A is 240 rpm counterclockwise and disk B i

mel-nik [20]

Answer:

a) αA = 4.35 rad/s²

αB = 1.84 rad/s²

b) t = 3.7 rad/s²

Explanation:

Given:

wA₀ = 240 rpm = 8π rad/s

wA₁ = 8π -αA*t₁

The angle in B is:

$\theta _{B} =4\pi =\frac{1}{2} \alpha _{B} t_{1}^{2} =\frac{1}{2} (\frac{r_{A} }{r_{B} } )^{3} \alpha _{A} t_{1}^{2}=\frac{1}{2} (\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}^{2}$

$\alpha _{A} =8\pi (\frac{0.2}{0.15} )^{3} =59.57rad$

$w_{B,1} =\alpha _{B} t_{1}=(\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}=0.422\alpha _{A} t_{1}$

The velocity at the contact point is equal to:

$v=r_{A} w_{A} =0.15*(8\pi -\alpha _{A} t_{1})=1.2\pi -0.15\alpha _{A} t_{1}$

$v=r_{B} w_{B} =0.2*(0.422\alpha _{A} t_{1})=0.0844\alpha _{A} t_{1}$

Matching both expressions:

$1.2\pi -0.15\alpha _{A} t_{1}=0.0844\alpha _{A} t_{1}\\\alpha _{A} t_{1}=16.09rad/s$

b) The time during which the disks slip is:

$t_{1} =\frac{\alpha _{A} t_{1}^{2}}{\alpha _{A} t_{1}} =\frac{59.574}{16.09} =3.7s$

a) The angular acceleration of each disk is

$\alpha _{A}=\frac{\alpha _{A} t_{1}}{t_{1} } =\frac{16.09}{3.7} =4.35rad/s^{2} (clockwise)$

$\alpha _{B}=(\frac{0.15}{0.2} )^{3} *4.35=1.84rad/s^{2} (clockwise)$

6 0

3 years ago

A hot-air balloon is accelerating upward under the influence of two forces, its weight and the buoyant force. for simplicity, co

Elenna [48]

Let V = the volume of the balloon
Force of gravity = V * ?hot * g downward
Buoyant force = V * ?cool * g upward
Net upward force F = V * ?cool * g - V * ?hot * g

F = V g (?cool - ?hot)

Mass of the balloon m = V ?hot

a = F/m = V g (?cool - ?hot)/(V ?hot)

a = g(?cool/?hot - 1)

a = 9.8(1.29/0.93 - 1)

a = 3.79 m/s^2

<span>Answer is 3.79 m/s^2</span>

4 0

3 years ago