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kotegsom [21]
3 years ago
11

A bicycle takes 8.0 seconds to accelerate at a constant rate from rest to a speed of 4.0 m/s. If the mass of the bicycle and rid

er together is 85 kg, what is the net force acting on the bicycle?
Physics
2 answers:
alexira [117]3 years ago
7 0

42.5 or 43 whichever answer you have on your test

Inessa05 [86]3 years ago
4 0

Acceleration = (change in speed) / (time for the change)
Acceleration = (4 m/s) / (8 seconds)
Acceleration = 0.5 m/s²

Force = (mass) x (acceleration)
Force = (85 kg) x (0.5 m/s²)
Force = 42.5 Newtons
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What is A rounded to the nearest tenth?
aniked [119]

Answer: B

Explanition:

7 0
3 years ago
A rectangular block has the density of 350g/cm3 the dimensions are 3.5 6.cm 2.5
LekaFEV [45]

Answer:

18375g

Explanation:

\boxed{density =  \frac{mass}{volume} }

∴ mass = density \times volume

Let's find the volume of the rectangular block.

Volume

= length ×breadth ×height

= 3.5 ×6 ×2.5

= 52.5cm³

Mass of the block

= 350(52.5)

= 18375g

7 0
3 years ago
Four power transistors, each dissipating 12 W, are mounted on a thin vertical aluminum plate 22 cm 3 22 cm in size. The heat gen
Sveta_85 [38]

Answer:

The temperature of the Aluminium plate 44.84⁰C

Explanation:

Number of transistors = 4

Since the heat dissipated by each transistor is 12W

Total heat dissipated, Q = 4 * 12 = 48 W

Q = 48 W

Cross sectional Area of the Aluminium plate, A = 2(l * b)

l = Length of the aluminium plate = 22 cm = 0.22 m

b = width of the aluminium plate = 22 cm = 0.22 m

A =2( 0.22 * 0.22 )

A = 0.0968 m²

From the heat balance equation, Q = hAΔT

h = 25 W/m²·K

A = 0.0968 m²

ΔT = T - T(air)

T(air) = 25°C

ΔT = T - 25°C

Q = 25 * 0.0968 * (  T - 25)

Q = 2.42 (T - 25)

Substitute Q = 48 into the equation above

48 =  2.42 (T - 25)

T - 25 = 19.84

T = 25 + 19.84

T = 44.84 ⁰C

6 0
3 years ago
What will happen if a car experiences a 300 N force to the right from the engine and a separate 150 N force due to friction and
gayaneshka [121]

Force applied on the car due to engine is given as

F_1 = 300 N towards right

Also there is a force on the car towards left due to air drag

F_2 = 150 N towards left

now the net force on the car will be given as

\vec F_{net} = \vec F_1 + \vec F_2

now we can say that since the two forces are here opposite in direction so here the vector sum of two forces will be the algebraic difference of the two forces.

So we can say

F_{net} = F_1 - F_2

F_{net} = 300 - 150

F_[net} = 150 N

So here net force on the car will be 150 N towards right and hence it will accelerate due to same force.

5 0
3 years ago
A test tube of length L and cross-sectional area A is submerged in water with the open end down so that the edge of the tube is
Margaret [11]

Answer:

if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.

Explanation:

The air in the tube can be considered an ideal gas,

           P V = nR T

In that case we have the tube in the air where the pressure is P1 = P_atm, then we introduce the tube to the water to a depth H

For pressure the open end of the tube is

         P₂ = P_atm + ρ g H

Let's write the gas equation for the colon

            P₁ V₁ = P₂ V₂

            P_atm V₁ = (P_atm + ρ g H) V₂

             V₂ = V₁    P_atm / (P_atm + ρ g h)

If the air obeys Boyle's law e; volume within the had must decrease due to the increase in pressure, if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.

The main assumption is that the temperature during the experiment does not change

6 0
3 years ago
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