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3 years ago

6

In physics class, Carrie learns that a force, F, is equal to the mass of an object, m, times its acceleration, a. She writes the

equation F=ma. Using this formula, what is the acceleration of an object with F=7.92 newtons and m=3.6 kilograms? Express your answer to the nearest tenth.

2 answers:

hjlf3 years ago

6 0

Answer:

your answer my good sir/lady is <u>C</u>

Explanation:

Phantasy [73]3 years ago

5 0

Answer:

2.2 m/s^2

Explanation:

Acceleration = Force / Mass

= 7.92 / 3.6 = 2.2m/s^2

Hope this help you :3

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Subduction occurs when:

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Explanation:

a plate with oceanic crust sinks beneath a plate with

continental crust is the answer

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4 years ago

NEED HELP ASAP!! Will give Brainliest.

olga2289 [7]

a. 25 joules of work is done by the object.

Explanation:

According to the work-energy theorem, the work done ON an object is equal to the change in kinetic energy of the object, therefore:

$W=K_f -K_i= \frac{1}{2}mv^2-\frac{1}{2}mu^2$

where

$K_f = \frac{1}{2}mv^2$ is the final kinetic energy of the car, with

m = 2 kg being the mass of the car

v = 0 is the final speed of the car (brought to rest)

$K_i = \frac{1}{2}mu^2$ is the initial kinetic energy of the car, with

u = 5 m/s being the initial speed of the car

Substituting numbers, we find:

$W=\frac{1}{2}(2)(0)^2 - \frac{1}{2}(2)(5)^2=-25 J$

The negative sign means that the work done ON the car is negative: this means therefore that the work has been actually done BY the car. In fact, the car has lost its kinetic energy: this means that it has done work on the surrounding, converting its kinetic energy into other forms of energy.

Learn more about work and kinetic energy:

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4 years ago

How might a theory relate to a model

Oliga [24]

A theory can be represented as a model

4 0

4 years ago

A luggage handler pulls a 20.0 kgkg suitcase up a ramp inclined at 32.0 ∘∘ above the horizontal by a force F⃗ F→ of magnitude 16

Nuetrik [128]

Answer:

a) W₁ = 8242.2 J, b) W₁ = 8242.2 J , c) W₃ = 0 , d) W₄ = -189.51 J ,

f) v = 27.24 m / s

Explanation:

a) Work is defined by

W = F. d = F d sin θ

where angle is between force and displacement

n this case the suitcase is going up and the outside F is parallel to the plane, so the angle is zero and the cosine is 1

W = F d

Let's calculate

W = 169 3.8

W₁ = 8242.2 J

b) the gravitational force is vertical so it has an angle with respect to the horizontal parallel to the plane of

θ’= 90 - θ

θ'= 90-32 = 58º

W = m g d thing θ ’

W = 20 9.8 3.8 thing (180 + tea ’) =

W = 744.8 cos (180 + 32)

W₂ = -631.6 J

c) The normal work, as it has 90º with respect to the displacement, its work is zero

W₃ = 0

d) the work of the friction force

Let's write Newton's second law the Y axis

N- Wy = 0

Cos 32 = Wy / W

N = W cos 32

The expression for friction force is

fr = μ N

fr = μ mg cos 32

fr = 0.300 20 9.8 cos (32)

fr = 49.87 N

The work of the friction force

W = fr d cos 180

W₄ = -49.87 3.8

W₄ = -189.51 J

E) The total work

W = W₁ + W₂ + W₃ + W₄

W = 8242.2- 631.6 + 0 -189.51

W_total = 7421.09 J

F) Usmeosel theorem of work and energy

W = ΔK

W = ΔK = ½ m v² - 0

v =√ 2W / m

v = √ (2 7421.09 / 20)

v = 27.24 m / s

5 0

4 years ago

Find the position vector of a particle that has the given acceleration and the specified initial velocity and position. a(t) = 1

kondor19780726 [428]

Answer:

Explanation:

Given

Acceleration $a(t)=14t\hat{i]+\sin (t)\hat{j}+\cos (2t)\hat{k}$ [/tex]

and $v(0)=\hat{i}$

$r(0)=\hat{j}$

we know $a=\frac{\mathrm{d} v}{\mathrm{d} t}$

$\int dv=\int adt$

$v(t)=\int (14t\hat{i}+\sin (t)\hat{j}+\cos (2t)\hat{k})dt$

$v(t)=7t^2\hat{i}-\cos t\hat{j}+\frac{\sin (2t)\hat{k}}{2}+c$

at $t=0$

$v(0)=0-1\cdot \hat{j}+0+c$

$c=\hat{i}+\hat{j}$

$v(t)=(7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2}$

and $\frac{\mathrm{d} r}{\mathrm{d} t}=v(t)$

$\int dr=\int vdt$

$r(t)=\int ((7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2})dt$

$r(t)=(\frac{7}{2}t^3+t)\hat{i}+(t-\sin (t))\hat{j}+\frac{1}{2}\times (-\frac{1}{2}\cos 2t)\hat{k}+c_2$

at $t=0$

$r(0)=\hat{j}$

$r(t)=(\frac{7}{3}t^3+t)\hat{i}+(1+t-\sin t)\hat{j}+\frac{1}{4}(1-\cos 2t)\hat{k}$

4 0

3 years ago