All categories

3 years ago

Which type of force is changing the object's speed and direction when a toy clown pops out of a jack-in-the-box?

Physics

2 answers:

pochemuha3 years ago

4 0

The correct answer is

B. push

To be more precise, the force that changes the speed and direction of the toy clown is the elastic force due to the spring: when the box is closed, the spring is compressed, so all the energy is stored in the elastic potential energy store of the spring. As the box opens, the spring is free to stretch, therefore it pushes the toy out of the box, transferring all the elastic potential energy into kinetic energy of motion of the toy clown.

Murrr4er [49]3 years ago

3 0

I think it would be d because the spring or whatever was pushing until it reached the farthest it could then it would pull down but idrk

You might be interested in

describe how your data supports the statement the rate of movement of dna fragments is inversely proportional to the log of thei

wlad13 [49]

Answer: Is there supposed to be something elses here because this question is unclear.

Explanation:

5 0

2 years ago

A diesel engine does not use spark plugs to ignite the fuel and air in the cylinders. Instead, the temperature required to ignit

Klio2033 [76]

Answer:

Compression ratio = 24.42

Explanation:

From Thermodynamic,

Adiabatic Equation ⇒ TV^{γ-1} = Constant.

⇒T₁V₁^{γ-1} = T₂V₂^{γ-1} ..................(1)

Where T₁= initial Temperature, V₁ = Initial Volume, T₂ = final Temperature, V₂ = Final Volume.

Given: T₁= 18°C we convert to Kelvin(K) by adding 273.

∴ T₁= 18°C + 273 = 291K

T₂ = 733° C also,we convert to Kelvin(K) by adding 273

∴ T₂ = 733° C +273 =1046K

γ = 7/5

∴ Rearranging equation(1), we have

(T₁/T₂) = (V₂/V₁)^{γ-1}................(2)

also rearranging equation(2) we have

(V₁/V₂)^{γ-1} = (T₂/T₁).

Where (V₁/V₂) = Compression ratio.

∴ (V₁/V₂)^{(7/5)-1} =( 1046/291)

simplifying the index in the equation

I.e (7/5)-1 = (7-5 )/5 = 2/5.

(V₁/V₂)^2/5 =(1046/291)^2/5

Multiplying the power on both side of the equation by 5/2.

∴(V₁/V₂)^(2/5)×(5/2) = (1046/291)^(5/2)

⇒ V₁/V₂= (1046/291)^2.5=( 3.59)^2.5

V₁/V₂ = 24.42.

∴ Compression ratio = 24.42

7 0

3 years ago

A flashlight beam strikes the surface of a pane of glass (n=1.56) at a 71° angle to the normal. What is the angle of refraction?

Aleks04 [339]

Answer:

The angle of refraction is 37°.

Explanation:

let n1 be the refractive index of glass and n2 = 1.0 be the refraction index of air, θ be the angle of incidence , ∅ be the angle of refraction.

then, according to Snell's law:

n1×sin(∅) = n2×sin(θ)

sin(∅) = n2×sin(θ)/n1

= (1.0)(sin(71°))/(1.56)

= 0.606101651

∅ = 37°

Therefore, the angle of refraction is 37°.

3 0

3 years ago

A small glass ball rubbed with silk gains a charge of +2.0 uc. the glass ball is placed 12 cm from a small charged rubber ball t

Gre4nikov [31]

The magnitude of the electric force between two obejcts with charge $q_1$ and $q_2$ is given by Coulomb's law:
$F= k_e \frac{q_1 q_2}{r^2}$
where
$k_e = 8.99 \cdot 10^9 N m^2 C^{-2}$ is the Coulomb's constant
and r is the distance between the two objects.

In our problem, the distance is $r=12 cm=0.12 m$ , while the magnitudes of the two charges are
$q_1 = 2.0 \mu C=2.0 \cdot 10^{-6}C$
$q_2 = 3.5 \mu C = 3.5 \cdot 10^{-6} C$
(we can neglect the sign of the second charge, since we are interested only in the magnitude of the force).

So, using the formula and the data of the problem, we find
$F=(8.99 \cdot 10^9 N m^2 C^{-2} ) \frac{(2.0 \cdot 10^{-6} C)(3.5 \cdot 10^{-6} C)}{(0.12 m)^2}=4.37 N$

4 0

3 years ago

A particle is projected at an angle 60 degrees to the horizontal with a speed of 20m/s. (i) calculate total time of flight of th

Kisachek [45]

Answer:

Time of flight=3.5 seconds

Speed at maximum height is 0

Explanation:

Φ=60°

initial velocity=u=20m/s

Acceleration due to gravity=g=9.8 m/s^2

Total time of flight=T

Final speed=v

question 1:

T=(2 x u x sinΦ)/g

T=(2 x 20 x sin60)/9.8

T=(2 x 20 x 0.8660)/9.8

T=34.64/9.8

T=3.5 seconds

Question 2

Speed at maximum height is 0

8 0

3 years ago