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3 years ago

What is the kind of energy that comes from movement?

Physics

2 answers:

Sedbober [7]3 years ago

8 0

The movement is motion energy

Brainiest plz

Sergio [31]3 years ago

4 0

Answer:

motion energy

Explanation:

motion wnergy is the sum of potential and kinetic energy

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How does friction increase and what do you call friction that is to strong that an applied force cannot overcome it?

Contact [7]

Well you see friction works by you moving in motion so why not try that.

6 0

3 years ago

According to Newton's first law of motion, which force is expected to cause a body to accelerate?

Nikitich [7]

Answer:

Unbalanced force -

Explanation:

Unbalanced force is expected to cause a body to accelerate.

4 0

3 years ago

Read 2 more answers

If the volume of a container increases, then the pressure will:

Elza [17]

d because the pressure particals won't have a small space to bump to

8 0

3 years ago

There is 50.0 mL of a gas at a temperature of 5°C. Its pressure is 760 mm Hg. What temperature is needed

tatuchka [14]

Answer:

41°C

Explanation:

Ideal gas law:

PV = nRT

where P is absolute pressure,

V is volume,

n is number of moles,

R is universal gas constant,

and T is absolute temperature.

n is held constant, so:

P₁V₁ / T₁ = P₂V₂ / T₂

Given:

P₁ = 760 mmHg

V₁ = 50.0 mL

T₁ = 5°C = 278 K

P₂ = 780 mmHg

V₂ = 55 mL

Find: T₂

(760 mmHg) (50.0 mL) / (278 K) = (780 mmHg) (55 mL) / T

T = 314 K

T = 41°C

8 0

3 years ago

The force on an object is F⃗ =−17j⃗ . For the vector v⃗ =2i⃗ +3j⃗ , find: (a) The component of F⃗ parallel to v⃗

Igoryamba

Answer:

(a) $\vec F_{\parallel} = -\frac{102}{13}\,i-\frac{103}{13}\,j$ , (b) $\vec F_{\perp} = \frac{102}{13}\,i -\frac{68}{13}\,j$ , (c) $W = -51$

Explanation:

The statement is incomplete:

The force on an object is $\vec F = -17\,j$ . For the vector $\vec v = 2\,i +3\,j$ . Find: (a) The component of $\vec F$ parallel to $\vec v$ , (b) The component of $\vec F$ perpendicular to $\vec v$ , and (c) The work $W$ , done by force $\vec F$ through displacement $\vec v$ .

(a) The component of $\vec F$ parallel to $\vec v$ is determined by the following expression:

$\vec F_{\parallel} = (\vec F \bullet \hat {v} )\cdot \hat{v}$

Where $\hat{v}$ is the unit vector of $\vec v$ , which is determined by the following expression:

$\hat{v} = \frac{\vec v}{\|\vec v \|}$

Where $\|\vec v\|$ is the norm of $\vec v$ , whose value can be found by Pythagorean Theorem.

Then, if $\vec F = -17\,j$ and $\vec v = 2\,i +3\,j$ , then:

$\|\vec v\| =\sqrt{2^{2}+3^{3}}$

$\|\vec v\|=\sqrt{13}$

$\hat{v} = \frac{1}{\sqrt{13}} \cdot(2\,i + 3\,j)$

$\hat{v} = \frac{2}{\sqrt{13}}\,i+ \frac{3}{\sqrt{13}}\,j$

$\vec F \bullet \hat{v} = (0)\cdot \left(\frac{2}{\sqrt{13}} \right)+(-17)\cdot \left(\frac{3}{\sqrt{13}} \right)$

$\vec F \bullet \hat{v} = -\frac{51}{\sqrt{13}}$

$\vec F_{\parallel} = \left(-\frac{51}{\sqrt{13}} \right)\cdot \left(\frac{2}{\sqrt{13}}\,i+\frac{3}{\sqrt{13}}\,j \right)$

$\vec F_{\parallel} = -\frac{102}{13}\,i-\frac{153}{13}\,j$

(b) Parallel and perpendicular components are orthogonal to each other and the perpendicular component can be found by using the following vectorial subtraction:

$\vec F_{\perp} = \vec F - \vec F_{\parallel}$

Given that $\vec F = -17\,j$ and $\vec F_{\parallel} = -\frac{102}{13}\,i-\frac{153}{13}\,j$ , the component of $\vec F$ perpendicular to $\vec v$ is:

$\vec F_{\perp} = -17\,j -\left(-\frac{102}{13}\,i-\frac{153}{13}\,j \right)$

$\vec F_{\perp} = \frac{102}{13}\,i + \left(\frac{153}{13}-17 \right)\,j$

$\vec F_{\perp} = \frac{102}{13}\,i -\frac{68}{13}\,j$

(c) The work done by $\vec F$ through displacement $\vec v$ is:

$W = \vec F \bullet \vec v$

$W = (0)\cdot (2)+(-17)\cdot (3)$

$W = -51$

8 0

3 years ago