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3 years ago

you ride a bike for 2 hours at 25 km/hr then 3 hours at 34 km/hr. what is your average velocity (in km/hr)?

Physics

1 answer:

mamaluj [8]3 years ago

4 0

<h3>Answer:</h3>

30.4 km/hr

<h3>Explanation:</h3>

<u>We are given</u>;

Speed in the first 2 hours as 25 km/hr
Speed in the next 3 hours as 34 km/hr

We are required to determine the average velocity in km/hr

To get the average velocity we divide total distance by total time.
Thus, we need to determine the total distance

Distance = Speed × time

Distance covered in the first 2 hours;

= 25 km/hr × 2 hours

= 50 km

Distance in the next 3 hours

= 34 km/hr × 3 hours

= 102 km

Therefore, total distance = 50 km + 102 km

= 152 km

Total time = 2 hrs + 3 hrs

= 5 hours

Therefore;

Average speed = 152 km ÷ 5 hours

= 30.4 km/hr

Thus, the average speed is 30.4 km/hr

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Nitroglycerin flows through a pipe of diameter 3.0 cm at 2.0 m/s. If the diameter narrows to 0.5 cm, what will the velocity be?

Korolek [52]

Answer:

72 m/s

Explanation:

D1 = 3 cm, v1 = 2 m/s

D2 = 0.5 cm,

Let the velocity at narrow end be v2.

By use of equation of continuity

A1 v1 = A2 v2

3.14 × 3 × 3 × 2 = 3.14 × 0.5 ×0.5 × v2

v2 = 72 m/s

8 0

3 years ago

A 5-m steel beam is lowered by means of two cables unwinding at the same speed from overhead cranes. As the beam approaches the

ArbitrLikvidat [17]

Answer:

a) The angular acceleration of the beam is 0.5 rad/s²CW (direction clockwise due the tangential acceleration is positive)

b) The acceleration of point A is 3.25 m/s²

The acceleration of point E is 0.75 m/s²

Explanation:

a) The relative acceleration of B with respect to D is equal:

$a_{B} =a_{D} +(a_{B/D} )_{n} +(a_{B/D} )_{t}$

Where

aB = absolute acceleration of point B = 2.5 j (m/s²)

aD = absolute acceleration of point D = 1.5 j (m/s²)

(aB/D)n = relative acceleration of point B respect to D (normal direction BD) = 0, no angular velocity of the beam

(aB/D)t = relative acceleration of point B respect to D (tangential direction BD)

$a_{B} =a_{D} +(a_{B/D} )_{t}$

$2.5j=1.5j +(a_{B/D} )_{t}\\(a_{B/D} )_{t}=j=1m/s^{2}$

We have that

(aB/D)t = BDα

Where α = acceleration of the beam

BDα = 1 m/s²

Where

BD = 2

$2\alpha =1\\\alpha =0.5rad/s^{2}CW$

b) The acceleration of point A is:

$a_{A} =a_{D} +(a_{A/D} )_{t}$

(aA/D)t = ADαj

$a_{A} =a_{D} +AD\alpha j\\a_{A}=1.5j+(3.5*0.5)j\\a_{A}=3.25jm/s^{2}$

The acceleration of point E is:

(aE/D)t = -EDαj

$a_{E} =a_{D} -ED\alpha j\\a_{E}=1.5j-(1.5*0.5)j\\a_{E}=0.75jm/s^{2}$

7 0

3 years ago

4. How do ordinary objects behave in microgravity?

Nikolay [14]

Answer:

Bubbles paused

Explanation:

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for instance, onto the International Space Station, gravity becomes negligible, and the laws of physics act differently than here on Earth

On Earth, the buoyancy of the air bubbles causes them to rise to the top together, creating a segregation between air and water. However, in microgravity, nothing forces the air bubbles to interact and thus rise together, Green said.

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Check out this app! It's millions of students helping each other get through their schoolwork. https://brainly.app.link/qpzV02Ma

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Explanation:

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3 years ago

A worker uses a cart to move a load of bricks a distance of 10m across a parking lot. if he pushes the cart with a constant forc

andriy [413]

Answer:

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