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soldi70 [24.7K]
2 years ago
13

Write a hypothesis about the effect of the fan speed on the acceleration of the cart. Use the "if . . . then . . . because . . .

” format and be sure to answer the lesson question: "How does an object's position and velocity change as the object accelerates?”
Physics
2 answers:
Hoochie [10]2 years ago
7 0

Answer:

If the fan speed increases, then the acceleration increases because a greater fan speed supplies more energy to move the cart.

Explanation:

Edge

Dimas [21]2 years ago
3 0

Answer:

As an object accelerates i.e., change it's velocity(either direction or speed), the position of the object depends on two factor; If the acceleration was direction based then it might have a zero displacement for eg: if it travels in circle. or it might have a net displacement if it travels in a straight line, quantitatively

where,

s = displacement

u = initial velocity

v = final velocity

a = acceleration

t = time

Now, for the hypothesis;

There is no direct relationship between fan speed and acceleration but anyways generally speaking if we do have a relationship that with more fan speed we have a larger displacement of air i.e., a more force i.e., greater acceleration

Thus, it can be said, well not exactly scientific, that with a greater fan speed there will be greater acceleration. if fan speed is increased then acceleration will be more.

:)

Explanation:

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vekshin1

Answer: D

Explanation:

5 0
2 years ago
In very cold weather, a significant mechanism for heat loss by the human body is energy expended in warming the air taken into t
Pie

Answer:

A) Q_a=74256\ J

B) Q=93562560\ J

Explanation:

Given:

  • temperature of air, T_a=-19+273=254\ K
  • temperature of lungs, T_l=37+273=310\ K
  • specific Heat exchanged from the lungs , c_l=0.47\ J.kg^{-1}.K^{-1}
  • specific heat of air, c_a=1020\ J.kg^{-1}.K^{-1}
  • mass of 1 L air, m'=1.3\ kg
  • breath rate, b=21\ breath.min^{-1}

A)

Now,

amount of heat needed to warm the air of lungs to the body temperature:

Q_a=m'.c_a.\Delta T

Q_a=1.3\times1020\times (310-254)

Q_a=74256\ J

B)

Amount of heat lost per hour:

<u>No. of breaths per hour:</u>

B=b.60

B=21\times 60

B=1260

<u>Now the total loss of energy in 1 hr.:</u>

Q=Q_a.B

Q=74256\times 1260

Q=93562560\ J

7 0
2 years ago
Consider a bird that flies at an average speed of 10.7 m/sm/s and releases energy from its body fat reserves at an average rate
Wittaler [7]

Answer:

455165.278 m

Explanation:

P = Power = 3.7 W

v = Velocity = 10.7 m/s

Amount of fat = 4 g

1 gram of fat provides about 9.40 (food) Calories

Energy given by 4 g of fat

E=4\times 9.4\times 4186\\\Rightarrow E=157393.6\ J

Time required to burn the fat

t=\dfrac{E}{P}\\\Rightarrow t=\dfrac{157393.6}{3.7}\\\Rightarrow t=42538.811\ s

Distance traveled by the bird

s=vt\\\Rightarrow s=10.7\times 42538.811\\\Rightarrow s=455165.2777\ m

The bird will fly 455165.278 m

4 0
2 years ago
Read 2 more answers
A 0.305 kg book rests at an angle against one side of a bookshelf. The magnitude and direction of the total force exerted on the
tankabanditka [31]

Answer

given,

F_L= 1.52\ N

\theta_L= 31^0

mass of book = 0.305 Kg

so, from the diagram attached  below

F_L cos {\theta_L} + F_b sin {\theta_b} = m g

1.52 times cos {31^0} + F_b sin {\theta_b} = 0.305 \times 9.8

F_b sin {\theta_b} = 2.989 -1.303

F_b sin {\theta_b} = 1.686

computing horizontal component

F_b cos {\theta_b} = F_L sin {\theta_L}

cos {\theta_b} = \dfrac{F_L sin {\theta_L}}{F_b}

cos {\theta_b} = \dfrac{1.52 \times sin {31^0}}{1.686}

cos {\theta_b} = 0.464

θ = 62.35°

5 0
2 years ago
A 7.5 nC point charge and a - 2.9 nC point charge are 3.2 cm apart. What is the electric field strength at the midpoint between
Oduvanchick [21]

Answer:

Net electric field, E_{net}=91406.24\ N/C

Explanation:

Given that,

Charge 1, q_1=7.5\ nC=7.5\times 10^{-9}\ C

Charge 2, q_2=-2.9\ nC=-2.9\times 10^{-9}\ C

distance, d = 3.2 cm = 0.032 m

Electric field due to charge 1 is given by :

E_1=\dfrac{kq_1}{r^2}

E_1=\dfrac{9\times 10^9\times 7.5\times 10^{-9}}{(0.032)^2}

E_1=65917.96\ N/C

Electric field due to charge 2 is given by :

E_2=\dfrac{kq_2}{r^2}

E_2=\dfrac{9\times 10^9\times 2.9\times 10^{-9}}{(0.032)^2}

E_2=25488.28\ N/C

The point charges have opposite charge. So, the net electric field is given by the sum of electric field due to both charges as :

E_{net}=E_1+E_2

E_{net}=65917.96+25488.28

E_{net}=91406.24\ N/C

So, the electric field strength at the midpoint between the two charges is 91406.24 N/C. Hence, this is the required solution.

3 0
2 years ago
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