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4 years ago

The force that every mass exerts on every other mass is called _____?

Physics

2 answers:

kirill115 [55]4 years ago

7 0

Weight or matter hope it helps

BlackZzzverrR [31]4 years ago

4 0

Gravity.
on Earth, the Earth's gravity is greater than all other gravity, so it seems like there's no gravity actually between objects. However, gravity is a force that exists between every mass.

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A man walks 30 m to the west than 5 m to the east in 45 seconds.

harkovskaia [24]

Answer:

35 m

0.56 m/s west

Explanation:

A) Total distance is the length of the path taken.

30 m + 5 m = 35 m

B) Velocity is displacement over time. Displacement is the difference between the final position and the initial position.

If west is -x, and east is +x, then:

Δx = -30 m + 5 m

Δx = -25 m

v = Δx / t

v = -25 m / 45 s

v = -0.56 m/s

v = 0.56 m/s west

4 0

3 years ago

A sled is at rest at the top of a 2 m high slope. The sled has a mass of 45 kg. The sled's potential energy isJ. (Formula: PE =

elixir [45]

Data:

h = 2m
m = 45 Kg
PE = ? (Joule)

Adopting, gravity (g) ≈ 9,8 m/s²

Formula: $PE_{grav} = mass * g * height$

Solving:

$PE_{grav} = mass * g * height$
$PE_{grav} = 45*9,8*2$
$PE_{grav} = 882J$

Answer:
The sled's potential energy is 882 Joules

8 0

3 years ago

Read 2 more answers

What is the time required for an object starting from rest.

sashaice [31]

Time required for what?

6 0

2 years ago

Ô tô chuyển động trên đường cong có bán kính R = 300(m). Nếu vận tốc ôtô tăng đều từ v1 =

kow [346]

Answer:

Mc ,12,4

Explanation:

7 0

3 years ago

A single Oreo cookie provides 53 kcal of energy. An athlete does an exercise that involves repeatedly lifting (without accelerat

Sever21 [200]

Answer:

Approximately $325$ (rounded down,) assuming that $g = 9.81\; {\rm N \cdot kg^{-1}}$ .

The number of repetitions would increase if efficiency increases.

Explanation:

Ensure that all quantities involved are in standard units:

Energy from the cookie (should be in joules, ${\rm J}$ ):

$\begin{aligned} & 53\; {\rm kCal} \times \frac{1\; {\rm kJ}}{4.184\; {\rm kCal}} \times \frac{1000\; {\rm J}}{1\; {\rm kJ}} \approx 2.551 \times 10^{5}\; {\rm J} \end{aligned}$ .

Height of the weight (should be in meters, ${\rm m}$ ):

$\begin{aligned} h &= 2\; {\rm dm} \times \frac{1\; {\rm m}}{10\; {\rm dm}} = 0.2\; {\rm m}\end{aligned}$ .

Energy required to lift the weight by $\Delta h = 0.2\; {\rm m}$ without acceleration:

$\begin{aligned} W &= m\, g\, \Delta h \\ &= 100\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times 0.2\; {\rm m} \\ &= 196\; {\rm N \cdot m} \\ &= 196\; {\rm J} \end{aligned}$ .

At an efficiency of $0.25$ , the actual amount of energy required to raise this weight to that height would be:

$\begin{aligned} \text{Energy Input} &= \frac{\text{Useful Work Output}}{\text{Efficiency}} \\ &= \frac{196\; {\rm J}}{0.25} \\ &=784\; {\rm J}\end{aligned}$ .

Divide $2.551 \times 10^{5}\; {\rm J}$ by $784\; {\rm J}$ to find the number of times this weight could be lifted up within that energy budget:

$\begin{aligned} \frac{2.551 \times 10^{5}\; {\rm J}}{784\; {\rm J}} &\approx 325 \end{aligned}$ .

Increasing the efficiency (the denominator) would reduce the amount of energy input required to achieve the same amount of useful work. Thus, the same energy budget would allow this weight to be lifted up for more times.

4 0

2 years ago