7) While having an existential crisis you lay on the floor to stare at the ceiling. The normal force you experience from the ground is likely equal to...

A The parallel component of the your weight
B Your weight
C The friction between the you and the ground
D Your mass

Answer:

options B,C,D are true

Explanation:

Answer:

walking to school

Explanation:

Driving a car to school

, and taking the bus to school​ both take up energy, unlike walking to school.

unless ur talking about energy, counting energy you produce and use to complete things, then it would be the 3rd one, taking the bus to school.

Answer:

Explanation:

for rolling motion down the plane acceleration is given by the following expression

a = g sinθ / (1 + k² / R²)

here k is radius of gyration and R is radius of the object rolling down .

for cylinder I = 1/2 m R²

so k² = R² / 2

k² / R² = 1/2

a = g sinθ /( 1 + 1 / 2 )

= 2 / 3 x  g sinθ

v = √ 2 a s

= √ (2 x  2 / 3 x  g sinθ s )

= √ (4  / 3 x  g h  )

= √ (4  / 3 x  g x .5  )

= √ 2g / 3

for sphere  I = 2/5  m R²

so k² = 2/5 R²

k² / R² = 2 / 5  

a = g sinθ / (1 + 2 / 5)  

= 5 / 7  x  g sinθ

v = √ 2 a s

= √ (2 x  5 / 7  x  g sinθ s )

= √ (10/7  x  g h  )

Given

√ (10/7  x  g h  ) = √ 2g / 3

10/7  x  g h  = 2g / 3

h = 14 / 30 m

= .47 m .

Answer:

jack has done more work pulling more weight and Jill has more power.

Explanation:

We know, [tex]1\ rpm = \dfrac{2\pi}{60} \ rad/s[/tex] .

[tex]64\ rpm\ is = \dfrac{2\pi}{60}\times 64\ rad/s\\\\= \dfrac{32\pi}{15}\ rad/s[/tex]

We know, kinetic energy is given by :

[tex]K.E = \dfrac{I\omega^2}{2}\\\\I = \dfrac{2(K.E)}{\omega^2}\\\\I = \dfrac{2\times 10^6}{\dfrac{32}{15}\times \pi}\\\\I = 298415.52 \ kg \ m^2[/tex]

Hence, this is the required solution.

Answer:

44 grams- 55 grams through the whole day. Probably about 14.6 grams per meal.

Explanation:

Answer:

2 thumbs (ISSA Guide)

Explanation:

Answer:

Explanation:

The momentum of an object can be found by using the formula

momentum = mass × velocity

From the question we have

momentum = 60 × 2

We have the final answer as

Hope this helps you

Answer:

the answer is true.......

Answer:

light refraction

Explanation:

Answer:

a)t  = 1,43 s

b) V = 10,49 m/s

c) V₀ₓ = 10,49 m/s   ;    V₀y = 14,01 m/s

d) Vf = 17,5 m/s

Explanation:

According to the problem statement

V₀ = V₀ₓ    and  V₀y = 0

And at the end of the movement t = ?  the distance y = 10 m

Therefore as

h = V₀y - (1/2)*g*t²

Vertical distance y = h = 10 = V₀y*t - 0,5 (-9,8)*t²

10 = 4,9*t²

t² = 10/4,9    ⇒  t² = 2,04 s

t  = 1,43 s

a) 1,43 s is the time of movement

b) V₀ = V₀ₓ        V₀y = 0     and  V₀ₓ = Vₓ     ( constant )

Just before touching the ground, the horizontal distance is

hd = 15 = Vₓ * t

Then  15 /1,43 = Vₓ = V₀ₓ

Vₓ = 10,49 m/s

Then initial speed is V = 10,49 m/s    since V₀y = 0

Vf² = Vₓ² + Vy²

Vyf = V₀y - g*t

Vyf =  0 - 9,8 *1,43

Vyf = - 14,01 m/s

And finally the speed when the projectile strike the ground is:

Vf² = Vₓ² + Vy²

Vf = √ (10,49)² + (14,01)²

Vf = 17,50 m/s

Answer:

Sensitivity of biosensor

The biosensor showed good linear correlation in the wide detection range of 0.001–2000 ng/mL with good sensitivity. In addition, it retained its biosensing property for seven days with high reproducibility

Explanation:

Answer:

The sun is the ultimate source of energy for all food chains. Through the process of photosynthesis, plants use light energy from the sun to make food energy. Energy flows, or is transferred through the system as one organism consumes another.

Answer:

s = 3.26 m

Explanation:

Given that,

Water leaves at a speed of 8 m/s

We need to find the maximum height that steelhead can jump. Let it can jump to a height of h.

At maximum height, final speed is equal to 0. We can use third equation of motion to find the maximum height.

[tex]v^2-u^2=2as[/tex]

a = -g

[tex]-u^2=-2gs\\\\s=\dfrac{u^2}{2g}\\\\s=\dfrac{(8)^2}{2\times 9.8}\\\\=3.26\ m[/tex]

Hence, the maximum height is 3.26 m.

Answer:

a. The bowling ball would have more kinetic energy because of its greater mass.

b. Potential energy

Explanation

a. Bowling ball has higher mass, self explanatory.

b. A high diver has lots of stored energy when they are on the diving platform. When they dive this stored energy helps make the splash when they hit the water. Stored energy is also called potential energy.

Answer:

It’s a

Explanation:

Don’t actually put that i needed the points mb

Answer:

I think it's most likely true.

Explanation:

any organism has the properties of a living thing, which includes cells, whether it has one cell or many

Answer:

False

Explanation:

An organism is a living thing that is a single-celled life form

Answer: Final temperature is 34.15°C.

Explanation: When two objects have different temperature, they will exchange heat energy until there is no more net energy transfer between them. At that state, the objects are in thermal equilibrium.

So, when in equilibrium, the total heat flow must be zero, i.e.:

[tex]Q_{1}+Q_{2}=0[/tex]

In our case, there will be a change in state of ice into water, so total heat flow will be:

[tex]m_{1}c_{1}(T_{f}-T_{i})+m_{2}c_{2}(T_{f}-T_{i})+mL=0[/tex]

where

m₁ is mass of ice

m₂ is mass of water

c₁ is specific heat of ice

c₂ is specific heat of water

[tex]T_{f}[/tex] is final temperature

[tex]T_{i}[/tex] is initial temperature

L is latent heat fusion

Temperature is in Kelvin so the transformation from Celsius to Kelvin:

For ice:

T = -15 + 273 = 258K

For water:

T = 48 + 273 = 321K

Solving:

[tex]21(2.09)(T_{f}-258)+158(4.186)(T_{f}-321)+21(333)=0[/tex]

[tex]43.89T_{f}-11323.62+661.4T_{f}-212305.55+6993=0[/tex]

[tex]705.3T_{f}=216636.17[/tex]

[tex]T_{f}=[/tex] 307.15K

In Celsius:

[tex]T_{f}=[/tex] 34.15°C

Final temperature of the system when in equilibrium is 34.15°C

Answer:

chemical symbol: Th

atomic number:90

protrons :90

neutrons:142

group#:4

period#: 9

Explanation:

you take the atomic weight (232.038)and subtract the atomic number to get (90) which is your neutrons

Answer:

Explanation:

The speed of the horse can be found by using the formula

[tex]v = \frac{p}{m} \\ [/tex]

p is the momentum

m is the mass

From the question we have

[tex]v = \frac{1200}{313} \\ = 3.83386..[/tex]

We have the final answer as

Hope this helps you

Answer:

The final velocity is 20 m/s.

Explanation:

Constant Acceleration Motion

It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.

Being a the constant acceleration, vo the initial speed, and t the time, the final speed can be calculated as follows:

[tex]v_f=v_o+at[/tex]

The provided data is: vo=10 m/s, [tex]a=5\ m/s^2[/tex], t=2 s. The final velocity is:

[tex]v_f=10~m/s+5\ m/s^2\cdot 2\ s[/tex]

[tex]v_f=20\ m/s[/tex]

The final velocity is 20 m/s.

Answer:

I = 0.5 A

Explanation:

Given that,

Average power of a light bulb, P = 60 W

The rms voltage supplied is 120 V

We need to find the value of rms current flowing through the light bulb. The relation between the power, voltage and current is given by :

P = VI

[tex]I=\dfrac{P}{V}\\\\I=\dfrac{60}{120}\\\\I=0.5\ A[/tex]

So, the rms value of current is 0.5 A.

Answer:

Ball 1 has a mass of 0.5 kilogram and an initial velocity of 1.00 meter/second. Ball 2 has a mass of 1.5 kg and an initial velocity of 0.00 meters/second.

Explanation:

Ball 1 has a mass of 0.5 kilogram and an initial velocity of 1.00 meter/second. Ball 2 has a mass of 1.5 kg and an initial velocity of 0.00 meters/second.

A collision is any situation in which two or more bodies quickly exert forces on one another. Despite the fact that the most common usage of the word "collision" refers to situations in which two or more objects clash violently, the scientific usage of the word makes no such assumptions.

The following are a few instances of physical encounters that scientists might classify as collisions. Legs of an insect are said to collide with a leaf when it falls on one.

Every contact of a cat's paws with the ground while it strides across a lawn is seen as a collision, as is every brush of its fur with a blade of grass.

Therefore, Ball 1 has a mass of 0.5 kilogram and an initial velocity of 1.00 meter/second. Ball 2 has a mass of 1.5 kg and an initial velocity of 0.00 meters/second.

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Answer:

Explanation:

Position of n the order fringe = n λ D / d

for n = 2

position = 2 λ D / d

λ = 710 nm , D = 1.1 m

d = .65 x 10⁻³

position 1 = 2 x 710 x 10⁻⁹ x 1.1 / .65 x 10⁻³

= 2403.07 x 10⁻⁶ m

= 2.403 x 10⁻³ m

= 2.403 mm .

For λ = 655 nm

position = 2 λ D / d

λ = 655 nm , D = 1.1 m

d = .65 x 10⁻³

position 2 = 2 x 655 x 10⁻⁹ x 1.1 / .65 x 10⁻³

= 2216.91 x 10⁻⁶ m

= 2.217 x 10⁻³ m

= 2.217 mm .

Difference between their position

= 2.403 - 2.217 = .186 mm .

Answer:

F = 288 [N]

Explanation:

To solve this problem we must use the following equation of kinematics and find the value of acceleration.

[tex]v_{f}^{2} =v_{o}^{2} +2*a*x[/tex]

where:

Vf =  final velocity = 6 [m/s]

Vo =  initial velocity = 0 (starting from rest)

a = acceleration [m/s²]

x = distance = 5 [m]

Now replacing, we have:

[tex](6)^{2}=0+(2*a*5)\\36=10*a\\a = 3.6 [m/s^{2}][/tex]

Since we already have the value of acceleration, we can use Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

ΣF = m*a

[tex]F =80*3.6\\F = 288 [N][/tex]

Answer:

Lifting a bag of groceries

Answer:

paddeling a canoe down a river :D or throwing a ball across a field

Explanation:

Answer:

a) h(g) = 358,53 m

b) t = 8,16 s

c) t(t) = 16,71 s

Explanation:

Equations for vertical shooting are:

Vf = V₀ -  g * t   ;     h  =  V₀*t -  (1/2)*g*t²  ;    Vf² = V₀² - 2*g*h

And at maximum heigt Vf = 0 then

0 = V₀ - g * t

t = V₀/g                     V₀  = 80 m/s      and   g = 9,8 m/s²

t  =  80 / 9,8   (s)

t = 8,16 s

Then 8,16 s is the time to get maximum height

If we plug t = 8,16 (s) in equation  h  =  V₀*t -  (1/2)*g*t²

we get:     h (max)  =  (80)*8,16 - 0,5*9,8*(8,16)² (m)

h (max) = 652,8  -  326,27 m

h (max) = 326,53 m

Then relative to ground that height becomes

h(g) = 326,53 + 32

h(g) = 358,53 m

In order to get the time the arrow is in the air we proceed as follows:

a) for the arrow to be at the launched point will take the same time that from the launched point to the maximum height, and after that we have to find out the time the arrow takes from 32 m down to the ground level

Then  

t(t) = 8,16 + 8,16 + tₓ     (2)

Where tₓ  is the time from 32 m height to ground

h  =  V₀*tₓ -  (1/2)*g*tₓ²  but since the arrow now is going down then we change the sign of the second term on the right side of the equation

32 = (80)*tₓ  +  0,5 * 9,8 * tₓ²     Note that when the arrow is at 32 m height the speed is again V₀ = 80 m/s

32 = 80*tₓ  + 4,9*tₓ²

A second-degree equation for tₓ, solving it

4,9*tₓ² + 80*tₓ - 32 = 0

t₁,₂ = -80 ± √ 6400 + 627,2 / 9,8

t₁,₂ =( - 80 ± 83,8 ) / 9,8

there is not a negative time therefore we dismiss such solution and

t₁ = 3,8 / 9,8

t₁ = 0,39 s

And

t(t) = 8,16 + 8,16 + 0,39  s

t(t) = 16,71 s

Answer:

[tex]-1.398\ \text{m/s}^2[/tex]

Decelerating or slowing down

Explanation:

F = Force = 50 N

[tex]\theta[/tex] = Angle force is being applied = [tex]42^{\circ}[/tex]

[tex]\mu[/tex] = Coefficient of friction = 0.64

m = Mass of wagon = 12 kg

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

Normal force is given by

[tex]N=mg-F\sin\theta[/tex]

Frictional force is given by

[tex]f=\mu N\\\Rightarrow f=\mu (mg-F\sin\theta)[/tex]

The force balance is given by

[tex]F\cos\theta-f=ma\\\Rightarrow \dfrac{F\cos\theta-\mu (mg-F\sin\theta)}{m}=a\\\Rightarrow a=\dfrac{50\times \cos42^{\circ}-0.64(12\times 9.81-50\times\sin42^{\circ})}{12}\\\Rightarrow a=-1.398\ \text{m/s}^2[/tex]

The acceleration of the wagon is [tex]-1.398\ \text{m/s}^2[/tex]. The negative sign indicates that the wagon is decelerating or slowing down.

The acceleration of the wagon is  [tex]-1.398 \;\rm m/s^{2}[/tex].

Given data:

The magnitude of pulling force is, F = 50 N.

The angle of inclination is, [tex]\theta = 42^{\circ}[/tex].

The mass of wagon wheel is, m = 12 kg.

Coefficient of friction between wagon and grass is, [tex]\mu =0.64[/tex].

The given problem is based on the concept of frictional force. The standard expression for the frictional force is,

[tex]f= \mu \times N[/tex]

Here, N is the normal force and its value is,

[tex]N=mg-Fsin \theta[/tex]

And the net force acting on wagon is,

[tex]F' = Fcos\theta -f\\\\ma = Fcos\theta -(\mu(mg-Fsin \theta))\\\\a = \dfrac{Fcos\theta -(\mu(mg-Fsin \theta))}{m}[/tex]

Here, a is the acceleration of wagon.

Solving as,

[tex]a = \dfrac{50 \times cos42 -(0.64(12 \times 9.8-(50 \times sin42)))}{12}\\\\a=-1.398 \;\rm m/s^{2}[/tex]

Thus, we can conclude that the acceleration of the wagon is  [tex]-1.398 \;\rm m/s^{2}[/tex].

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