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3 years ago

What pulls all objects in the universe, including the moon and earth and the su

1 answer:

6 0

It's called gravity, it attract the sun toward the gravitational pull making everything circulate. I don't really know how to explain it though.

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The constant forces F1 = 8 + 29 + 32 N and F2 = 48 - 59 - 22 N act together on a particle during a displacement from the point A

steposvetlana [31]

Answer:

- 600 J

Explanation:

A (20, 15, 0 ) m

B (0, 0, 7) m

$\overrightarrow{F_{1}}=8\widehat{i}+29\widehat{j}+32\widehat{k}$

$\overrightarrow{F_{2}}=48\widehat{i}-59\widehat{j}-22\widehat{k}$

Net force

$\overrightarrow{F}=\overrightarrow{F_{1}}+\overrightarrow{F_{2}}$

$\overrightarrow{F}}=\left ( 8+48 \right )\widehat{i}+\left ( 29-59 \right )\widehat{j}+\left ( 32-22 \right )\widehat{k}$

$\overrightarrow{F}}=56\widehat{i}-30\widehat{j}+10\widehat{k}$

$\overrightarrow{S}=\overrightarrow{OB}-\overrightarrow{OA}$

$\overrightarrow{S}=\left ( 0-20 \right )\widehat{i}+\left ( 0-15 \right )\widehat{j}+\left ( 7-0 \right )\widehat{k}$

$\overrightarrow{S}=-20\widehat{i}-15\widehat{j}+7\widehat{k}$

Work done is defined as

$W = \overrightarrow{F}.\overrightarrow{S}$

$W = \left ( 56\widehat{i}-30\widehat{j}+10\widehat{k} \right ).\left (-20\widehat{i}-15\widehat{j}+7\widehat{k} \right )$

W = -1120 + 450 + 70

W = - 600 J

3 0

3 years ago

A 250–g piece of gold is at 19 °C. 5.192 kJ of energy is added to it by heat. The specific heat of gold is 129 J/(kg·°C). Calcul

algol [13]

Answer:

A. DT is given by Q= MCs DT

m = mass of the substances

Cs= is it's specific heat capacity

Ck= Q

Mk ×DTk

=250 × 9 × 5

129

=Dt = 180.1085271

answer is 180degree C.

Explanation:

B. = 25×10 ×100

1.082

=2500

1.082

= 23105.360 g/kj.

7 0

3 years ago

How do you take mold off a bigmac

ollegr [7]

Answer:

ksjsnjsnsjsjjsjdjd dsjsjv sbsbbsbshdi udhudushsjjd dydyshyehehwheuwuwe dydyshyehehwheuwuwe what's he dudj

6 0

3 years ago

What is the maximum height a 95 W motor may vertically lift a 120 N weight in 6.2 s?

leva [86]

Answer:

a. 4.9 m

Explanation:

To solve this problem we must take into account that power is defined as the relationship between the work and the time in which the work is done.

P = W/t

where:

P = power = 95 [W] (units of watts)

W = work [J] (units of Joules)

t = time = 6.2 [s]

We can clear the work from the previous equation.

W = P*t

W = 95*6.2 = 589 [J]

Now we know that the work is defined by the product of the force by the distance, therefore we can express the work done with the following equation.

W = F*d

where:

F = force = 120 [N] (units of Newtons)

d = distance [m]

d = W/F

d = 589/120

d = 4.9 [m]

4 0

3 years ago

An exoplanet has three times the mass and one-fourth the radius of the Earth. Find the acceleration due to gravity on its surfac

Harlamova29_29 [7]

Answer:

b. 48.0 g.

Explanation:

Given;

mass of the exoplanet, $M_p = 3M_e$

radius of the exoplanet, $r_p = \frac{1}{4} r_e$

The acceleration due to gravity of the planet is calculated as;

$g_p = \frac{GM_p}{r_p^2} \\\\for \ Earth's \ surface\\\\g = \frac{GM_e}{r_e^2} \\\\G = \frac{gr_e^2}{M_e} = \frac{g_pr_p^2}{M_p} \\\\\frac{gr_e^2}{M_e} = \frac{g_p(\frac{r_e}{4}) ^2}{3M_e} \\\\\frac{gr_e^2}{M_e} = \frac{g_pr_e ^2}{16\times 3M_e} \\\\g = \frac{g_p}{48} \\\\g_p = 48 \ g$

Therefore, the correct option is b. 48.0 g

5 0

2 years ago