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DedPeter [7]
2 years ago
11

.

title="25 {?}^{?} \times \frac{?}{?} " alt="25 {?}^{?} \times \frac{?}{?} " align="absmiddle" class="latex-formula">
​
Physics
2 answers:
blondinia [14]2 years ago
7 0
What are u asking exactly ?
mr_godi [17]2 years ago
5 0

Answer:

what is your exact question.

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a cyclist coasting down a 5.0 ◦ incline at a constant speed of 6.0 km/h because of air resistance. If the total mass of the bicy
Dvinal [7]

Answer:

F_{net}= 85.41\ N

Explanation:

mass of the bicycle + cyclist = 50 kg

constant speed = 6 km/h

a cyclist coasting down a 5.0° incline

the downward velocity is constant, so net acceleration must be zero

the air drag must be equal to gravitational force downward along the ramp

F_a = mg sin \theta  

now for upward motion

F_{net} = mg sin \theta + air\ drag

F_{net} = mg sin \theta + mg sin \theta

F_{net} = 2 mg sin \theta

F_{net} = 2\times 50 \times 9.8 sin 5^0

F_{net}= 85.41\ N

3 0
3 years ago
Please Help! Im dumb.
Rus_ich [418]
The answers false I believe
8 0
3 years ago
Read 2 more answers
Which statement best compares a scientific theory and an opinion?
Furkat [3]

Answer: c opinion is something ppl just suggest but a theory needs proof before it is confirmed

8 0
2 years ago
Which of the following are vector quantities?
pshichka [43]

Answer:

Vector Quantity: A physical quantity is said to be a vector quantity when it has both magnitude and direction. The scalar quantities are distance, mass, time, volume, density, speed, temperature, and energy, The vector quantities are weight, velocity, acceleration, and force.

Explanation:

Mark me brainleist PLZZZZ

8 0
3 years ago
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To test the performance of its tires, a car travels along a perfectly flat (no banking) circular track of radius 130 m. The car
marysya [2.9K]

Answer:

0.739

Explanation:

If we treat the four tire as single body then

W ( weight of the tyre ) =  mass × acceleration due to gravity (g)

the body has a tangential acceleration = dv/dt = 5.22 m/s², also the body has centripetal acceleration to the center = v² / r

where v is speed 25.6 m/s and r is the radius of the circle

centripetal acceleration = (25.6 m/s)² / 130 = 5.041 m/s²

net acceleration of the body = √ (tangential acceleration² + centripetal acceleration²) = √ (5.22² + 5.041²) = 7.2567 m/s²

coefficient of static friction between the tires and the road = frictional force / force of normal

frictional force = m × net acceleration / m×g

where force of normal = weight of the body in opposite direction

coefficient of static friction = (7.2567 × m) / (9.81 × m)

coefficient of static friction = 0.739

4 0
2 years ago
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