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3 years ago

Help please, this is unit about work, power, force in physics.

Physics

1 answer:

bearhunter [10]3 years ago

5 0

1 to 4 5 throu the 7 to 9 and then others are obviously to the 6 and 8

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You walk exactly 250 steps North, turn around, and then walk exactly 400 steps South. How far are you from your starting

german

Answer:

150 steps south

Explanation:

250 north 250 back to start then continue south for remainder of 400 steps. 150 south

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3 years ago

8750 J of heat are applied to a piece of aluminium causing a 56C increase in its

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Answer:146.8983 grams

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3 years ago

Globalization is the process of

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Answer: Globalization is the process of interaction and integration among people, companies, and governments worldwide.

Explanation:

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3 years ago

Read 2 more answers

A 40-kg box is being pushed along a horizontal smooth surface. The pushing force is 15 n directed at an angle of 15Â° below the

Kobotan [32]

Answer:

Acceleration of the crate is 0.362 m/s^2.

Explanation:

Given:

Mass of the box, m = 40 kg

Applied force, F = 15 N

Angle at which the force is applied, $(\theta)$ = 15°

We have to find the magnitude of the acceleration.

Let the acceleration be "a".

FBD is attached with where we can see the horizontal and vertical component of force.

⇒ $F_x=Fcos(\theta)$ and ⇒ $F_y=Fsin (\theta)$

⇒ $F_x=15cos(15)$ ⇒ $F_y=15sin (15)$

⇒ Applying concept of forces.

⇒ $\sum F_x=F_n_e_t =F-f$

⇒ $F_n_e_t =F-f$

⇒ $ma =F-f$ <em> ...Newtons second law Fnet = ma</em>

⇒ $a =\frac{F-f}{m}$

⇒ Plugging the values.

⇒ $a =\frac{15cos(15)-0}{40}$ <em>...f is the friction which is zero here.</em>

⇒ $a =\frac{14.48}{40}$

⇒ $a=0.362\ ms^-^2$

Magnitude of the acceleration of the crate is 0.362 m/s^2.

4 0

3 years ago

A bicycle racer inflates their tires to 7.1 atm on a warm autumn afternoon when temperatures reached 27 °C. By morning the tempe

natulia [17]

Answer:

The required pressure is 6.4866 atm.

Explanation:

The given data : -

In the afternoon.

Initial pressure of tire ( p₁ ) = 7 atm = 7 * 101.325 Kpa = 709.275 Kpa

Initial temperature ( T₁ ) = 27°C = (27 + 273) K = 300 K

In the morning .

Final temperature ( T₂ ) = 5°C = ( 5 + 273 ) K = 278 K

Given that volume remains constant.

To find final pressure ( p₂ ).

Applying the ideal gas equation.

p * v = m * R * T

$\frac{p}{T} = constant$

$\frac{p_{1} }{T_{1} } = \frac{p_{2} }{T_{2} }$

$p_{2} = \frac{T_{2} }{T_{1} } *p_{1}$

$p_{2} = \frac{278}{300} * 709.275$ = 657.2615 Kpa = 6.486 atm

8 0

3 years ago