A football player kicks a 0.41-kg football initially at rest; and the ball flies through the air. If the kicker's foot was in contact with the ball for 0.051 s and the ball's initial speed after the collision is 21 m/s, what was the magnitude of the average force on the football?

Answer:

(a)

[tex]\lambda _{m}=9.332 \times 10^{-6}m[/tex]

(b)

[tex]\lambda _{m}=1.632 \times 10^{-6}m[/tex]

(c) [tex]\lambda _{m}=4.988 \times 10^{-7}m[/tex]

Explanation:

According to the Wein's displacement law

[tex]\lambda _{m}\times T = b[/tex]

Where, T be the absolute temperature and b is the Wein's displacement constant.

b = 2.898 x 10^-3 m-K

(a) T = 37°C = 37 + 273 = 310 K

[tex]\lambda _{m}=\frac{b}{T}[/tex]

[tex]\lambda _{m}=\frac{2.893\times 10^{-3}}{310}[/tex]

[tex]\lambda _{m}=9.332 \times 10^{-6}m[/tex]

(b) T = 1500°C = 1500 + 273 = 1773 K

[tex]\lambda _{m}=\frac{b}{T}[/tex]

[tex]\lambda _{m}=\frac{2.893\times 10^{-3}}{1773}[/tex]

[tex]\lambda _{m}=1.632 \times 10^{-6}m[/tex]

(c) T = 5800 K

[tex]\lambda _{m}=\frac{b}{T}[/tex]

[tex]\lambda _{m}=\frac{2.893\times 10^{-3}}{5800}[/tex]

[tex]\lambda _{m}=4.988 \times 10^{-7}m[/tex]

Answer:

Part a)

[tex]U = 13 J[/tex]

Part b)

[tex]v = 2.28 m/s[/tex]

Part c)

[tex]v = 177.66 m/s[/tex]

Part d)

[tex]W = 1012.7 J[/tex]

Part e)

[tex]v = 2.1 m/s[/tex]

Part f)

[tex]E = 1037.2 J[/tex]

Explanation:

Part a)

As we know that the maximum angle deflected by the pendulum is

[tex]\theta = 38^o[/tex]

so the maximum height reached by the pendulum is given as

[tex]h = L(1 - cos\theta)[/tex]

so we will have

[tex]h = L(1 - cos38)[/tex]

[tex]h = 1.25(1 - cos38)[/tex]

[tex]h = 0.265 m[/tex]

now gravitational potential energy of the pendulum is given as

[tex]U = mgh[/tex]

[tex]U = 5(9.81)(0.265)[/tex]

[tex]U = 13 J[/tex]

Part b)

As we know that there is no energy loss while moving upwards after being stuck

so here we can use mechanical energy conservation law

so we have

[tex]mgh = \frac{1}{2}mv^2[/tex]

[tex]v = \sqrt{2gh}[/tex]

[tex]v = \sqrt{2(9.81)(0.265)}[/tex]

[tex]v = 2.28 m/s[/tex]

Part c)

now by momentum conservation we can say

[tex]mv = (M + m) v_f[/tex]

[tex]0.065 v = (5 + 0.065)2.28[/tex]

[tex]v = 177.66 m/s[/tex]

Part d)

Work done by the bullet is equal to the change in kinetic energy of the system

so we have

[tex]W = \frac{1}{2}mv^2 - \frac{1}{2}(m + M)v_f^2[/tex]

[tex]W = \frac{1}{2}(0.065)(177.66)^2 - \frac{1}{2}(5 + 0.065)2.28^2[/tex]

[tex]W = 1012.7 J[/tex]

Part e)

recoil speed of the gun can be calculated by momentum conservation

so we will have

[tex]0 = mv_1 + Mv_2[/tex]

[tex]0 = 0.065(177.6) + 5.50 v[/tex]

[tex]v = 2.1 m/s[/tex]

Part f)

Total energy released in the process of shooting of gun

[tex]E = \frac{1}{2}Mv^2 + \frac{1}{2}mv_1^2[/tex]

[tex]E = \frac{1}{2}(5.50)(2.1^2) + \frac{1}{2}(0.065)(177.6^2)[/tex]

[tex]E = 1037.2 J[/tex]

Answer:

20 kg

Explanation:

Kinetic energy=½*Mass * velocity²

4000= ½* m*20²

8000=400m

m=8000/400

m=20 kg

Answer:

a≅3.33

Explanation:

Answer:

The acceleration of the block down the incline is 3.23 m/s²

Explanation:

Fk = frictional force

m = mass

g = acceleration due to gravity

F1 = force in direction of motion

theeta = inclination angle

For Force on inclined plane

F1 = m*g*sin(theeta)

F1 = (1.2)*(9.8)*sin(30)

F1 = 5.88-N

now,

Fnet = F1 - Fk

Fnet = 5.88 - 2

Fnet = 3.88-N

now for acceleration,

Fnet = m*a

a = Fnet / m

a = 3.88 / 1.2

a = 3.23 m/s²

Answer:

    [tex]x= 32.544[/tex]

Explanation:

given,

mean weight of bag (μ) = 32

standard deviation (σ) = 0.32

percentage of bag heavier = 4.5%

weight of the bag less than 4.5 % = 100 - 4.5

                                                        = 95.5%

we have to determine the z- value according to 95.5% or 0.955

using z-table

     z-value = 1.70

now, using formula

       [tex]Z = \dfrac{x-\mu}{\sigma}[/tex]

       [tex]1.70 = \dfrac{x-32}{0.32}[/tex]

       [tex]x-32 = 1.70\times {0.32}[/tex]

      [tex]x= 32.544[/tex]

The most a bag of baby carrots can weigh and not need to be repackaged is approximately 32.53 ounces.

To determine the most a bag of baby carrots can weigh and not need to be repackaged, we need to find the weight corresponding to the upper 4.5% of the normal distribution.

The weights are normally distributed with a mean of 32 ounces and a standard deviation (σ) of 0.32 ounces.

We first find the z-score that corresponds to the top 4.5% of the distribution.

Using a z-table or calculator,

Next, we use the z-score formula:

Solving for X (the weight we need):

Multiplying both sides by 0.32:

Adding 32 to both sides:

Thus, the most a bag of baby carrots can weigh and not need to be repackaged is approximately 32.53 ounces.

Answer:

W = 8085.064 J

Explanation:

given,

mass of the cheerleaders partner = 37.4 Kg

distance above which she was lift = 0.817 m

acceleration due to gravity = 9.8 m/s²

number of time she was picked = 27 times

work he done = ?

now,                                    

Work done will be equal to the potential energy into number times she was lifted.                                          

Work done =  N m g h                    

              W = 27 x 37.4 x 9.8 x 0.817

              W = 8085.064 J                                  

work done by his partner is equal to W = 8085.064 J

Answer:

A)

<17, - 1, 0>

B)

0 Ns

C)

<17, - 1, 0>

D)

<4, - 5, 0>

Explanation:

A)

[tex]p_{A,i}[/tex] = initial momentum of object A = 15 i - 8 j + 0 k

[tex]p_{B,i}[/tex] = initial momentum of object B = 2 i + 7 j + 0 k

Total initial momentum of the system is given as the sum of initial momenta of A and B , hence

[tex]p_{i}[/tex] = [tex]p_{A,i}[/tex] + [tex]p_{B,i}[/tex]

[tex]p_{i}[/tex] = (15 i - 8 j + 0 k) + (2 i + 7 j + 0 k)

[tex]p_{i}[/tex] = 17 i - j + 0 k

B)

[tex]F_{ext}[/tex] = Net external force on the two objects = 0 N

[tex]t[/tex] = duration of the collision

[tex]I[/tex] = Impulse

Impulse is given as

[tex]I = F_{ext} t \\I = (0) t \\I = 0 Ns[/tex]

C)

[tex]p_{f}[/tex] = final momentum of the system

we know that the Impulse is nothing but change in momentum of the system of objects, hence

[tex]I = p_{f} - p_{i}\\0 = p_{f} - p_{i}\\p_{f} = p_{i}[/tex]

[tex]p_{f}[/tex] = 17 i - j + 0 k

D)

[tex]p_{A,f}[/tex] = final momentum of object A = 13 i + 4 j + 0 k

[tex]p_{B,f}[/tex] = final momentum of object B = ?

Total final momentum of the system is given as

[tex]p_{f} = p_{A,f} + p_{B,f}[/tex]

17 i - j + 0 k = ( 13 i + 4 j + 0 k ) + [tex]p_{B,f}[/tex]

[tex]p_{B,f}[/tex] = 4 i - 5 j + 0 k

So final momentum of the object is <4, - 5, 0>

The total initial momentum of this system just before the collision is 17 i - j + 0 k.

This is defined as a term which quantifies the overall effect of a force acting over time.

Initial momentum of A = 15 i - 8 j + 0 k

initial momentum of B = 2 i + 7 j + 0 k

Total = sum of initial momentum of A and B

= (15 i - 8 j + 0 k) + (2 i + 7 j + 0 k)

= 17 i - j + 0 k

Impulse = Ft where F is force and t is time

Net external force on the two objects = 0 N

Impulse = (0)t

              = 0Ns

Impulse = change in momentum

= 17 i - j + 0 k

Final momentum of object A = 13 i + 4 j + 0 k

Final momentum of object B = ?

Total final momentum = Final momentum of object A + Final momentum of object B

17 i - j + 0 k = ( 13 i + 4 j + 0 k ) + Final momentum of object B

= 4 i - 5 j + 0 k

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Final answer:

The acceleration of the crate on the frictionless ramp is 2.39 m/s², while the acceleration on the ramp with friction is 0.49 m/s².

Explanation:

To find the acceleration of the crate, we first need to determine the force component parallel to the incline. This can be done using the formula F_parallel = F * sin(theta), where F is the applied force and theta is the angle of the ramp. Substituting the values, we have F_parallel = 7.0 N * sin(20°) = 2.39 N. The acceleration can then be found using the equation a = F_parallel / m, where m is the mass of the crate. Substituting the values, we have a = 2.39 N / 1.0 kg = 2.39 m/s².

If the ramp has a friction force of 1.9 N, it will oppose the motion of the crate. In this case, we need to subtract the friction force from the parallel force to find the effective force. The effective force, Feff, is given by Feff = F_parallel - friction force. Substituting the values, we have Feff = 2.39 N - 1.9 N = 0.49 N. The acceleration can then be found using the equation a = Feff / m. Substituting the values, we have a = 0.49 N / 1.0 kg = 0.49 m/s².

The two possible frequencies could be ; 642 Hz or 712 Hz

Explanation:

For beats; f₁-f₂=fb where f₁=frequency of the first guitar, f₂=frequency of the second guitar, fb =frequency of beat

In this case, the two possibilities are;

f₁-f₂=fb or f₂-f₁=fb

f₁=677 Hz  and fb=35 Hz

then;

f₁-x=fb

677-x=35

x=677-35=642 Hz

or

x-f₁=fb

x-677=35

x=35+677 =712 Hz

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The object initially had potential energy of 108,410 Joules when placed 6.5 meters higher. Since it maintains the same speed, this energy was used to overcome friction. Thus, the object lost 108,410 Joules to friction.

To answer your question, we first need to calculate the initial potential energy of the object when it was 6.5 m higher. Potential energy (PE) is given by PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height. So, the initial potential energy is PE = (1700 kg)(9.8 m/s²)(6.5 m) = 108,410 Joules.

Since the object reaches the top of the loop with the same velocity in both scenarios, we can say that its kinetic energy remained the same. This implies all of the initial potential energy went into overcoming friction. Therefore, the object lost 108,410 Joules of energy to friction.

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The object lost 111,230 joules of energy to friction during its ascent.

The question is asking how much energy, in joules, the object lost to friction if it starts 6.5 m higher than the previous part and still reaches the top of the loop with the same velocity. To calculate this, we need to find the change in gravitational potential energy, which is given by the formula: ΔPE = mgh.

The mass of the object is 1700 kg, and the change in height is 6.5 m. Substituting these values into the equation, we have:

ΔPE = (1700 kg)(9.8 m/s^2)(6.5 m) = 111,230 J.

Therefore, the object lost 111,230 joules of energy to friction during its ascent.

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Answer:

(a) Potential energy of the child is converted into the kinetic energy at the bottom off the slide and a part of which is lost into friction generating heat between the contact surfaces.

(b) [tex]U=668.16\ J[/tex]

Explanation:

Given:

(a)

∴The initial potential energy of the child is converted into the kinetic energy at the bottom off the slide and a part of which is lost into friction generating heat between the contact surfaces.

Initial potential energy:

[tex]PE=m.g.h[/tex]

[tex]PE=24\times 9.8\times 3.3[/tex]

[tex]PE=776.16\ J[/tex]

Kinetic energy at the bottom of the slide:

[tex]KE=\frac{1}{2} m.v^2[/tex]

[tex]KE= 0.5\times 24\times 3^2[/tex]

[tex]KE= 108\ J[/tex]

(b)

Now, the difference in the potential and kinetic energy is the total change in the thermal energy of the slide and the seat of her pants.

This can be given as:

[tex]U=PE-KE[/tex]

[tex]U=776.16-108[/tex]

[tex]U=668.16\ J[/tex]

Answer:The gravitational field on Saturn can be calculated by the following formula;

Explanation:

Answer:

Glycosidic Bonds

Explanation:

When two adjacent mono-saccharide units link to form di-saccharides and this repeated chain link forms poly-saccharide. These reactions are called dehydration or condensation reactions because upon the formation of the bond the water molecule is released in the form of a byproduct. Glycosidic bonds are covalent bonds that link ring-shaped sugar molecules to other molecules.

Answer:

No change in the specific heat.

Explanation:

Specific heat is an intrinsic property that it has not depend upon on the amount of substance or energy added, it depends upon the material.

So, when twice the amount of energy is transferred, specific heat of the material does not change rather the energy that is twice in the amount to 1 kg of that material cause it to warm 2.0° C.

Answer:

B:the changes in the characteristics within a population that lead to survival.

Explanation:

Answer:

B

Explanation:

study Island

Answer:

1.5 W/kg, 3.47 W/kg

6 dogs

Explanation:

[tex]m_d[/tex] = Mass of dog = 38 kg

[tex]v_d[/tex] = Pulling speed of dog = 2.2 m/s

[tex]F_d[/tex] = Force on sled = 60 N

[tex]P_h[/tex] = Power of horse = 750 W

[tex]m_h[/tex] = Mass of horse = 500 kg

Power is given by

[tex]P_d=F_dv_d\\\Rightarrow P_d=60\times 2.2\\\Rightarrow P_d=132\ W[/tex]

Specific power is given by

[tex]P_{sh}=\frac{P_h}{m_h}\\\Rightarrow P_{sh}=\frac{750}{500}\\\Rightarrow P_{sh}=1.5\ W/kg[/tex]

Specific power of horse is 1.5 W/kg

[tex]P_{sd}=\frac{P_d}{m_d}\\\Rightarrow P_{sd}=\frac{132}{38}\\\Rightarrow P_{sh}=3.47\ W/kg[/tex]

Specific power of dog is 3.47 W/kg

We have the relation

[tex]nP_d=P_h\\\Rightarrow n=\frac{P_h}{P_d}\\\Rightarrow n=\frac{750}{132}=5.68\ dogs\approx 6\ dogs[/tex]

The number of dogs is 6

(1) The specific power of the horse is 1.5 W/kg and the specific power of the dog is 3.5 W/kg.

(2) The minimum number of dogs needed to provide the same power as one horse is 6 dogs.

The specific power of the horse is calculated as follows;

Cp = P/m

Cp = 750/500

Cp = 1.5 W/kg

The specific power of the dog is calculated as follows;

Cp = P/m

Cp = (Fv)/m
Cp = (60 x 2.2)/38

Cp = 3.5 W/kg

n(Fv) = 750

n = (750)/(60 x 2.2)

n = 6 dogs

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The acceleration at the bottom of the loop is 34m/[tex]\bold{s^2}[/tex]

Explanation:

Given:  

Mass=52kg

Force=1750N

To calculate:

The acceleration at the bottom of the loop

Formula:

Force=Mass x Acceleration

1750=52 x Acceleration

1750/52=Acceleration

Therefore acceleration at the bottom of the loop is 34m/[tex]s^2[/tex]

Roller coasters are mainly based upon acceleration theory they have two types of acceleration one is at the top of the loop and the other is at the bottom of the loop.

Then the net forces and the values are given. In many problems the roller coaster concept is included and it gives another level of clarity to the problems including the net forces

Answer: Minimum global acc when you substitute t = 0 in equation 2

Acc = 23.52m/s^2.

Global max acc = 35.595684524m/s^2

Explanation:

(t)= 0.001490833t^3-.08389t^2+23.52t+7.3(in feet per second). ...equation 1.

Equation 1 is the model showing the velocity pathway..

To derive the equation of acceleration.

We differentiate equation 1

DV(t)/Dt = acc. = .004472499t^2 - 0.16778t + 23.52...equ 2

Minimum global acc when you substitute t = 0 in equation 2

Acc = 23.52m/s^2.

Maximum global acc when you substitute the = 74s

Acc = 24.491404524 -12.41572 +23.52

Global max acc = 35.595684524m/s^2

To find the global maximum and minimum values of the acceleration of a rocket, we compute the second derivative of the velocity function, which is the acceleration function. We then evaluate the acceleration function at the given time points and at any critical points within this period.

To find the global maximum and minimum values of the acceleration of a rocket, we first need to determine the function for the acceleration, a(t), which is the second derivative of the given velocity function, v(t). The given velocity function v(t) = 0.001490833t^3 - 0.08389t^2 + 23.52t + 7.3. The first derivative, v'(t), gives us the acceleration function and can be computed as v'(t) = 0.004472499t^2 - 0.16778t + 23.52. Computing the second derivative, v''(t), we get a(t) = 0.008944998t - 0.16778.

To find the maximum and minimum values of this acceleration within the given time frame (0 to 74 sec), we evaluate the function at these time points and at any critical points within this interval. The critical points are obtained by solving the equation a'(t) = 0, which yield the value for which the acceleration is maximum or minimum.

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Answer:

The velocity of the student and skateboard together [tex]=1.82\ ms^{-1}[/tex]

Explanation:

Given:

Mass of student [tex]m_1=71\ kg[/tex]

Mass of skateboard [tex]m_2=2.8\ kg[/tex]

Distance between student and skateboard [tex]d=2.75\ m[/tex]

Acceleration of student [tex]a=0.65\ ms^{-2}[/tex]

Finding velocity [tex]v_1[/tex] of the student  before jumping on skateboard

Using equation of motion

[tex]v_1^2=v_0^2+2ad[/tex]

here [tex]v_0[/tex] represents the initial velocity of the student which is [tex]=0[/tex] as he starts from rest.

So,

[tex]v_1^2=0^2+2(0.65)(2.75)[/tex]

[tex]v_1^2=3.575[/tex]

Taking square root both sides:

[tex]\sqrt{v_1^2}=\sqrt[1.7875}[/tex]

∴ [tex]v_1=1.89[/tex]

Finding velocity [tex]v[/tex] of student and skateboard.

Using law of conservation of momentum.

[tex]m_1v_1+m_2v_2=(m_1+m_2)v[/tex]

Where [tex]v_2[/tex] is initial velocity of skateboard which is [tex]=0[/tex] as it is at rest.

Plugging in values.

[tex]71(1.89)+(2.8)(0)=(71+2.8)\ v[/tex]

[tex]134.19=73.8\ v[/tex]

Dividing both sides by [tex]73.8[/tex]

[tex]\frac{134.19}{73.8}=\frac{73.8\ v}{73.8}[/tex]

∴ [tex]v=1.82[/tex]

The velocity of the student and skateboard together [tex]=1.82\ ms^{-1}[/tex]

Answer:

6.62 m

About 44.14 grapefruits lined in a single line would be the length of the diameter of star Aldebaran

Explanation:

First we know,

Diameter of the Sun - 1.39 million km (Ds)

Diameter of Aldbaran - 61.40 million km (Da)

Thus,

[tex]\frac{Da}{Ds}[/tex] =  [tex]\frac{61.40}{1.39}[/tex] ≈ 44.14

We know that in this scale the size of the sun is that of a grapefruit.

The average diameter of a grapefruit is about 15cm

Thus  Ds in this scale is 15cm

Thus Da in this scale,

Da = Ds *44.14

Da = 0.15 * 44.14 ≈ 6.62 m

Thus the diameter of Aldebaran in this scale is 6.62 m

Answer:

h should become four times.

Explanation:

The speed at the bottom depends only on the vertical height difference from the top ... if there is no friction ... then the path you take from top to bottom doesn't matter in any way, only the difference in height between the two

mgh at top = 1/2 mv^2 at bottom, so that v=sqrt[2gh]

You see, the final velocity, v, depends on the square root of h, so double v, quadruple h, or four times higher

Answer:

maximum amplitude  = 0.13 m

Explanation:

Given that

Time period T= 0.74 s

acceleration of gravity g= 10 m/s²

We know that time period of simple harmonic motion given as

[tex]T=\dfrac{2\pi}{\omega}[/tex]

[tex]0.74=\dfrac{2\pi}{\omega}[/tex]

ω = 8.48 rad/s

ω=angular frequency

Lets take amplitude = A

The maximum acceleration given as

a= ω² A

The maximum acceleration should be equal to g ,then block does not separate

a= ω² A

10= 8.48² A

A=0.13 m

maximum amplitude  = 0.13 m

The maximum amplitude of the motion for which the block does not separate from the plate is : 0.14 m

Given data :

period ( T ) = 0.74 secs

acceleration due to gravity ( g ) = 9.8 m/s62

Time period of simple harmonic motion ( T ) = [tex]\frac{2\pi }{w}[/tex]

First step : solve for w

w = 2π / T

   = 2π / 0.74

   = 8.49 rad/sec

Next step : determine the maximum amplitude ( A )

a = w² A

where ;  a = maximum acceleration = 9.8 , w = 8.49

therefore

A = a / w²

   = 9.8 / ( 8.49 )²

   = 0.136  ≈ 0.14 m

Hence we can conclude that The maximum amplitude of the motion for which the block does not separate from the plate is : 0.14 m.

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Answer:

1. [tex]v=14.2259\ m.s^{-1}[/tex]

2. [tex]F_T=25.8924\ N[/tex]

3. [tex]\lambda=29.6373\ m[/tex]

Explanation:

Given:

1.

[tex]\rm Speed\ of\ wave\ pulse=Length\ of\ slinky\div time\ taken\ by\ the\ wave\ to\ travel[/tex]

[tex]v=\frac{6.8}{0.478}[/tex]

[tex]v=14.2259\ m.s^{-1}[/tex]

2.

Now, we find the linear mass density of the slinky.

[tex]\mu=\frac{m}{L}[/tex]

[tex]\mu=\frac{0.87}{6.8}\ kg.m^{-1}[/tex]

We have the relation involving the tension force as:

[tex]v=\sqrt{\frac{F_T}{\mu} }[/tex]

[tex]14.2259=\sqrt{\frac{F_T}{\frac{0.87}{6.8}} }[/tex]

[tex]202.3774=F_T\times \frac{6.8}{0.87}[/tex]

[tex]F_T=25.8924\ N[/tex]

3.

We have the relation for wavelength as:

[tex]\lambda=\frac{v}{f}[/tex]

[tex]\lambda=\frac{14.2259}{0.48}[/tex]

[tex]\lambda=29.6373\ m[/tex]

Impulse : 7.5.10⁻¹⁰N.s

Momentum is the product of the mass of an object and its velocity

Whereas impulse is defined is the product of the force with the time interval when the force is acting on the object.

the pollen grains to be ejected out of the flower in 0.30 ms and acceleration of 2.5 × 10⁴ m/s², then

Δt = 0.3 ms = 3.10⁻⁴ s

a = 2.5 × 10⁴ m/s²

mass of pollen: 1.0 × 10⁻⁷g = 1.0 x 10⁻¹⁰ kg

then:

A force that works according to Newton's second law:

[tex]\rm F=1.0\times 10^{-10}\times 2.5\times 10^4\\\\F=2.5\times 10^{-6}\:N[/tex]

Impulse (I) :

[tex]\rm I=2.5 \times 10^{-6}\times 3\times 10^{-4}\\\\I=7.5\times 10^{-10}N.s[/tex]

Collisions between 2 objects  

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displacement of A skateboarder

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The distance of the elevator  

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Final answer:

To calculate the impulse delivered to a pollen grain, the mass is converted to kilograms and then multiplied with the given acceleration to find the force. This force is then multiplied by the time over which it acts to obtain the impulse, which is 75 in units of 10⁻¹⁰ kgm/s.

Explanation:

The question asks us to calculate the impulse delivered to a pollen grain by the forceful opening of a bunchberry plant flower. To find the impulse (I), which is the change in momentum, we can use the formula I = Ft, where F is the force and t is the time that the force is applied. Alternatively, since the question provides the acceleration (a) and initial time, we can utilize the link between force, mass, and acceleration (F = ma) and then apply that result to find the impulse using the time.

Firstly, we need to convert the mass of the pollen grain to kilograms (kg) which is the standard unit of mass in physics equations:
1.0 × 10⁻⁷ g = 1.0 × 10⁻¹⁰ kg.
Now, we calculate the force on the pollen grain:
F = ma = (1.0 × 10⁻¹⁰ kg) × (2.5 × 10⁴ m/s²) = 2.5 × 10⁻⁵ N.
Next, with force and the time of application known, we can find the impulse:
I = Ft = (2.5 × 10⁻⁵ N) × (0.30 × 10⁻³ s) = 7.5 × 10⁻⁹ kg·m/s.
In terms of 10⁻¹⁰ kg·m/s, the impulse is 75.

The equation for the speed of a satellite in a circular orbit around the Earth depends on mass. Which mass?

a. The mass of the sun

b. The mass of the satellite

c. The mass of the Earth

The equation for the speed of a satellite orbiting in a circular path around the earth depends upon the mass of Earth.

Option c

Any particular body performing circular motion has a centripetal force in picture. In this case of a satellite revolving in a circular orbit around the earth, the necessary centripetal force is provided by the gravitational force between the satellite and earth. Hence [tex]F_{G} = F_{C}[/tex].

Gravitational force between Earth and Satellite: [tex]F_{G} = \frac{G \times M_e \times M_s}{R^2}[/tex]

Centripetal force of Satellite :[tex]F_C = \frac{M_s \times V^2}{R}[/tex]

Where G = Gravitational Constant

[tex]M_e[/tex]= Mass of Earth

[tex]M_s[/tex]= Mass of satellite

R= Radius of satellite’s circular orbit

V = Speed of satellite

Equating  [tex]F_G = F_C[/tex], we get  

Speed of Satellite [tex]V =\frac{\sqrt{G \times M_e}}{R}[/tex]

Thus the speed of satellite depends only on the mass of Earth.

Answer:

b. isothermal

Explanation:

When a thermodynamic process occurs and the temperature remains constant, it is said to be an isothermal process.

The prefix "iso" means "equal", so the word isothermal indicates that the temperature remains equal in the process, and that is what happens in the situation described, the volume is varied (compressed) while maintaining the sample of nitrogen gas remains at a constant temperature.

Answer:

1.19cm^3 of glycerine

Explanation:

Let Vo= 150cm^3 for both aluminum and glycerine, using expansion formula:

Volume of spill glycerine = change in volume of glycerine - change in volume of aluminum

Volume of glycerine = coefficient of volume expansion of glycerine * Vo* change in temperature - coefficient of volume expansion of Aluminum*Vo* change temperature

coefficient of volume expansion of aluminum = coefficient of linear expansion of aluminum*3 = 23*10^-6 * 3 = 0.69*10^-4 oC^-1

Change in temperature = 41-23 = 18oC

Volume of glycerine that spill = (5.1*10^-4) - (0.69*10^-4) (150*18) = 4.41*10^-4*2700 = 1.19cm3

Answer:

The rise in temperature is [tex]43.98^{\circ}C[/tex]

Solution:

As per the question:

Mass of hammer, M = 1.30 kg

Speed of hammer, v = 7.3 m/s

Mass of iron, [tex]m_{i} = 14\ g[/tex]

No. of blows, n = 8

Specific heat of iron, [tex]C_{i} = 450\ J/kg.^{\circ}C = 0.45\ J/g^{\circ}C[/tex]

Now,

To calculate the temperature rise:

Transfer of energy in a blow = Change in the Kinetic energy

[tex]\Delta KE = \frac{1}{2}Mv^{2} = \frac{1}{2}\times 1.30\times 7.3^{2} = 34.64\ J[/tex]

For 8 such blows:

[tex]\Delta KE = n\Delta KE = 8\times 34.64\ = 277.12 J[/tex]

Now, we know that:

[tex]Q = m_{i}C_{i}\Delta T[/tex]

[tex]\Delta T= \frac{\Delta KE}{m_{i}C_{i}}[/tex]

[tex]\Delta T= \frac{277.12}{14\times 0.45} = 43.98^{\circ}C[/tex]

Answer:

429 J  joules of heat are needed to raise the temperature of 5.00 g

Explanation:

given data

specific heat capacity = 2.20 j/g-k

methane = 5.00 g

temperature range = 36.0°c to 75.0°c

to find out

How many joules of heat are needed to raise the temperature of 5.00 g

solution

we will apply here formula for heat that needed to raise the temperature

Q = m × S × Δt     ..............................1

m is mass of methane and  S is specific heat capacity and Δt is change in temperature

put here value we get

Q = m × S × Δt  

Q = 5 × 2.2 × ( 75 - 36 )

Q = 5 × 2.2 × 39

Q =  429 J

so  429 J  joules of heat are needed to raise the temperature of 5.00 g

The total force on the spring is the sum of the weight of the 5-kg object and the additional force due to the elevator's upward acceleration. This force leads to a stretch in the spring, calculated using Hooke's law, of 0.064 m or 6.4 cm.

To solve this problem we'll use Hooke's law (F = -kx), where F is the force, k is the spring constant (in our case, 802 N/m), and x is the displacement or stretch of the spring.

Here, the force isn't just from the weight of the 5.0-kg object (mg), but also from the additional force due to the upward acceleration of the elevator (ma). By adding these two forces together, we can find the total force on the spring (F = mg + ma).

First, let's calculate the weight of the 5-kg object: F_weight = mg = 5.0 kg * 9.81 m/s^2 = 49.05 N.

Next, let's calculate the additional force due to the elevator's acceleration: F_acceleration = ma = 5.0 kg * 0.41 m/s^2 = 2.05 N.

Adding these two forces together: F_total = F_weight + F_acceleration = 49.05 N + 2.05 N = 51.1 N.

Now, we can use Hooke's law to find the stretch of the spring x = F_total / k = 51.1 N / 802 N/m = 0.064 m, or 6.4 cm when converted to centimeters.

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The spring stretches by approximately 0.020 meters relative to its unstrained length when the elevator is accelerating upward at 0.41 m/s^2.

The amount by which the spring stretches can be determined using Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring. In this case, the force exerted by the spring is equal to the weight of the object attached to it. The equation for this can be written as F = kx, where F is the force, k is the spring constant, and x is the displacement. Rearranging this equation, we can solve for x:

x = F / k

Plugging in the values given in the question, we have:

x = (m * a) / k

Substituting the values m = 5.0 kg, a = 0.41 m/s^2, and k = 802 N/m:

x = (5.0 kg * 0.41 m/s^2) / 802 N/m

x ≈ 0.020 m

Therefore, the spring stretches by approximately 0.020 meters relative to its unstrained length when the elevator is accelerating upward at 0.41 m/s^2.

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