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3 years ago

How can Newton's third law describe the forces affecting a rocket as it

Physics

2 answers:

Vitek1552 [10]3 years ago

6 0

Answer: the answer is C

Explanation: I just took the test

Alexxandr [17]3 years ago

6 0

Answer:

Explanation:

The force the rocket exerts on the air is in the direction opposite to

the force that the air exerts on the rocket.

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Answer <br> A)<br> B <br> C <br> please

babymother [125]

Answer:

Explanation:

brash this so hard

5 0

1 year ago

A crate lies on a plane tilted at an angle θ = 22.5 ∘ to the horizontal, with μk = 0.19.

ValentinkaMS [17]

A) $2.03 m/s^2$

Let's start by writing the equation of the forces along the directions parallel and perpendicular to the incline:

Parallel:

$mg sin \theta - \mu_k R = ma$ (1)

where

m is the mass

g = 9.8 m/s^2 the acceleration of gravity

$\theta=22.5^{\circ}$

$\mu_k = 0.19$ is the coefficient of friction

R is the normal reaction

a is the acceleration

Perpendicular:

$R-mg cos \theta =0$ (2)

From (2) we find

$R=mg cos \theta$

And substituting into (1)

$mg sin \theta - \mu_k mg cos \theta = ma$

Solving for a,

$a=g sin \theta - \mu_k g cos \theta = (9.8)(sin 22.5)-(0.19)(9.8)(cos 22.5)=2.03 m/s^2$

B) 5.94 m/s

We can solve this part by using the suvat equation

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

Here we have

v = ?

u = 0 (it starts from rest)

$a=2.03 m/s^2$

s = 8.70 m

Solving for v,

$v=\sqrt{u^2+2as}=\sqrt{0+2(2.03)(8.70)}=5.94 m/s$

6 0

3 years ago

Calculate the magnitude of the acceleration of the box if you push on the box with a constant force 170.0 N that is parallel to

Degger [83]

The acceleration of the box up the ramp is 9.65 m/s².

<h3>What is the magnitude of acceleration of the box?</h3>

The magnitude of the acceleration of the box is calculated by applying Newton's second law of motion as shown below;

F(net) = ma

where;

m is the mass of the box
a is the acceleration of the box

The net force on the box is calculated as follows;

F(net) = F - Ff

F(net) = F - μmgcosθ

where;

θ is the inclination of the plane
μ is coefficient of friction

F(net) = 170 - (0.3 x 15 x 9.8 x cos55)

F(net) = 144.7

The acceleration of the box is calculated as;

a = F(net) / m

a = (144.7) / (15)

a = 9.65 m/s²

Thus, the acceleration of the box up the ramp is 9.65 m/s².

Learn more about acceleration here: brainly.com/question/14344386

#SPJ4

8 0

10 months ago

A capacitor is made from two hollow, coaxial, iron cylinders, one inside the other. The inner cylinder is negatively charged and

Musya8 [376]

Answer: 3.12 * 10^12 F ( 3.12 pF)

Explanation: To calculate this capacitor of two hollow, coaxial, iron cylinders, we have to determine the potental differente between them and afeter that to use C=Q/ΔV

The electric field in th eregion rinner<r<router

By using the Gaussian law

∫E*ds=Q inside/εo

E*2*π*rinner^2*L= Q /εo

E=Q/(2*π*εo*r^2)

[Vab]=\int\limits^a_b {E} \, dr

where a and b are the inner and outer radii.

Then we have:

ΔV= 2*k*(Q/L)* ln (b/a)

replacing the values and using that C=Q/ΔV

we have:

C= L/(2*k*ln(b/a)=0.17/(2*9*10^9*3.023)=3.12 pF

5 0

2 years ago

Let’s look at a radio-controlled model car. Suppose that at time t1=2.0st1=2.0s the car has components of velocity vx=1.0m/svx=1

Nonamiya [84]

Answer:

$a_x=6\ \text{m/s}^2$ and $a_y=0\ \text{m/s}^2$

Magnitude of accleration is $6\ \text{m/s}^2$ and the direction is $0^{\circ}$

Explanation:

$t_1=2\ \text{s}$

$v_x=1\ \text{m/s}$

$v_y=3\ \text{m/s}$

$t_2=2.5\ \text{s}$

$v_x=4\ \text{m/s}$

$v_y=3\ \text{m/s}$

Average acceleration in the different axes

$a_x=\dfrac{\Delta v_x}{\Delta t}\\\Rightarrow a_x=\dfrac{4-1}{2.5-2}\\\Rightarrow a_x=6\ \text{m/s}^2$

$a_y=\dfrac{\Delta v_y}{\Delta t}\\\Rightarrow a_y=\dfrac{3-3}{2.5-2}\\\Rightarrow a_y=0\ \text{m/s}^2$

The components of the acceleration is $a_x=6\ \text{m/s}^2$ and $a_y=0\ \text{m/s}^2$

The magnitude of acceleration

$a=\sqrt{a_x^2+a_y^2}\\\Rightarrow a=\sqrt{6^2+0^2}\\\Rightarrow a=6\ \text{m/s}^2$

Direction

$\theta=\tan^{-1}\dfrac{a_y}{a_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{0}{6}\\\Rightarrow \theta=0^{\circ}$

The magnitude of accleration is $6\ \text{m/s}^2$ and the direction is $0^{\circ}$ .

7 0

2 years ago