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bija089 [108]
3 years ago
7

A force of 10N is required to stretch a spring from 20cm to 25cm. What is the spring constant in N/m2 Be careful of unit

Physics
1 answer:
kupik [55]3 years ago
5 0

Answer:

C) 40 N/m

Explanation:

If we ASSUME that the spring is un-stretched at the zero cm position

k = F/Δx = 10/0.25 = 40 N/m

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An object weighs 32 newtons. What is its mass if a gravitometer indicates that g = 8.25 m/s2?
Nata [24]

Answer:

3.88kg

Explanation:

From the question given, we obtained the following data:

W = 32N

g = 8.25 m/s2

M =?

W = M x g

M = W/g

M = 32/8.25

M = 3.88Kg

The mass of the object is 3.88kg

3 0
4 years ago
When jumping straight down, you can be seriously injured if you land stiff-legged. One way to avoid injury is to bend your knees
nexus9112 [7]

(a) The man comes to a halt in 1.99 ms = 1.99 × 10⁻³ s, so his average acceleration slowing him from 6.33 m/s to a rest is

<em>a</em> = (6.33 m/s - 0) / (1.99 × 10⁻³ s) ≈ 3180 m/s²

so that the average net force on the man during the landing is

<em>F</em> = (70.8 kg) <em>a</em> ≈ 225,000 N

i.e. with magnitude 225,000 N.

(b) With knees bent, the man has an average acceleration of

<em>a</em> = (6.33 m/s - 0) / (0.147 s) ≈ 43.1 m/s²

and hence an average net force of

<em>F</em> = (70.8 kg) <em>a</em> ≈ 3050 N

(c) The net force on the man is

∑ <em>F</em> = <em>n</em> - <em>w</em> = <em>m a</em>

where

<em>n</em> = magnitude of the normal force, i.e. the force of the ground pushing up on the man

<em>w</em> = the man's weight, <em>m g</em> ≈ 694 N

<em>m</em> = the man's mass, 70.8 kg

<em>g</em> = mag. of the acceleration due to gravity, 9.80 m/s²

<em>a</em> = the man's acceleration

Using the acceleration in part (b), we have

<em>n</em> = <em>m g</em> + <em>m a</em> = <em>m</em> (<em>g</em> + <em>a</em>)

<em>n</em> = (70.8 kg) (9.80 m/s² + 43.1 m/s²) ≈ 3740 N

4 0
3 years ago
Starting from rest, a 2.3x10-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exerts
Harman [31]

Answer:

3.13 m/s

Explanation:

From the question,

Since the flea spring started from rest,

Ek = W................... Equation 1

Where Ek = Kinetic Energy of the flea spring, W = work done on the flea spring.

But,

Ek = 1/2mv²............ Equation 2

Where m = mass of the flea spring, v = flea's speed when it leaves the ground.

substitute equation 2 into equation 1

1/2mv² = W.................... Equation 3

make v the subject of the equation

v = √(2W/m)................. Equation 4

Given: W = 3.6×10⁻⁴ J, m = 2.3×10⁻⁴ kg

Substitute into equation 4

v = √[2×3.6×10⁻⁴ )/2.3×10⁻⁴]

v = 7.2/2.3

v = 3.13 m/s

Hence the flea's speed when it leaves the ground  = 3.13 m/s

4 0
3 years ago
In 2017, the company SpaceX became the first private company to send supplies to the International Space Station with a reusable
pav-90 [236]

Answer:

Approximately 3.98\; \rm m \cdot s^{-2}.

Assumption: air resistance on the rocket is negligible. Take g = \rm 9.81\; m \cdot s^{-2}.

Explanation:

By Newton's Second Law of Motion, the acceleration of the rocket is proportional to the net force on it.

\displaystyle \text{Acceleration} = \frac{\text{Net Force}}{\text{Mass}}.

Note that in this case, the uppercase letter \rm M in the units stands for "mega-", which is the same as 10^6 times the unit that follows. For example, \rm 1\; Mg = 10^6\; g, while \rm 1\; MN = 10^6\; N.

Convert the mass of the rocket and the thrust of its engines to SI standard units:

  • The standard unit for mass is kilograms: \displaystyle m = \rm 552\; Mg = 552 \times 10^6\; g \times \frac{1\; \rm kg}{10^3\; g}  = 552 \times 10^3 \; kg.
  • The standard for forces (including thrust) is Newtons: \text{Thrust} = \rm 7.61 \; MN = 7.61 \times 10^6\; N.

At launch, the velocity of the rocket would be pretty low. Hence, compared to thrust and weight, the air resistance on the rocket would be pretty negligible. The two main forces that contribute to the net force of the rocket would be:

  • Thrust (which is supposed to go upwards), and
  • Weight (downwards due to gravity.)

The thrust on the rocket is already known to be \rm 7.61 \times 10^6\; N. Since the rocket is quite close to the ground, the gravitational acceleration on it should be approximately 9.81\; \rm m \cdot s^{-2} = 9.81 \; N \cdot kg^{-1}. Hence, the weight on the rocket would be approximately 9.81\; \rm N \cdot kg^{-1} \times 552 \times 10^3\; kg = 5.41412\times 10^6\; N.

The magnitude of the net force on the rocket would be

\begin{aligned}&\text{Thrust} - \text{Weight} \\ &= 7.61 \times 10^6\; \rm N - 5.41412\times 10^6\; N \\ &\approx 2.19 \times 10^6\; \rm N\end{aligned}.

Apply the formula \displaystyle \text{Acceleration} = \frac{\text{Net Force}}{\text{Mass}} to find the net force on the rocket. To make sure that the output (acceleration) is in SI units (meters-per-second,) make sure that the inputs (net force and mass) are also in SI units (Newtons for net force and kilograms for mass.)

\begin{aligned}\displaystyle &\text{Acceleration} \\ &= \frac{\text{Net Force}}{\text{Mass}} \\ &= \frac{2.19 \times 10^6\; \rm N}{552 \times 10^3\; \rm kg}  \\ &\approx \rm 3.98\; \rm m \cdot s^{-2}\end{aligned}.

6 0
3 years ago
Plss help quick!!! Which component of the galaxy is the arrow pointing to?
uranmaximum [27]
I think it’s dust sorry if it wrong
7 0
3 years ago
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