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2 years ago

Define the term “force”.

Physics

2 answers:

romanna [79]2 years ago

6 0

The Jedi have the force. it flows through all things. Joking it will change the motion of an object

Aleonysh [2.5K]2 years ago

3 0

Energy that is applied to an object.

--TheOneandOnly003

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You are sitting on a deck of your house surrounded by oak trees. You hear the sound of an acorn hitting the deck. You wonder if

Black_prince [1.1K]

Answer: 96N

Explanation:

To calculate the velocity of the impact On the persons head, we have

h = gt²/2

14 = 9.81t²/2

t² = 28/9.8

t² = 2.86

t = 1.69s

V = u + at

V = 0 + 9.81*1.69

V = 16.58m/s

a(average) = (v1² + v2²) /2Δy

a(average) = 16.58² + 0)/2 * 0.005

a(average) = 274.8964/0.01

a(average) = 27489.64m/s²

Using newton's second law of motion,

F(average) = m * a(average)

F(average) = 0.0035 * 27489.64

F(average) = 96.21N

Therefore the force needed by the acorn to do much damage starts from 96N

8 0

2 years ago

In this first example of constant accelerated motion, we will simply consider a car that is initially traveling along a straight

algol13

Answer:

$v_{f} =25m/s$

Explanation:

Kinematics equation for constant acceleration:

$v_{f} =v_{o} + at=15+2*5=25m/s$

4 0

2 years ago

What is the wavelength of a wave that has a speed of 350 meters/second and a frequency of 140 hertz?In meters

Leto [7]

Apply:
wavelength = speed/frequency
= 350 m/s : 140 Hz = 2.5 m.

3 0

2 years ago

Read 2 more answers

A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the ma

raketka [301]

Let, the maximum height covered by projectile be $\sf{H_m}$

$\purple{ \longrightarrow \bf{h_m = \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} }}$

Projectile is thrown with a velocity = v
Angle of projection = θ

Velocity of projectile at a height half of the maximum height covered be $\sf{v_0}$

$\qquad$ ______________________________

Then –

$\qquad$ $\pink{ \longrightarrow \bf{ \dfrac{h_m}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }}$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} \times \dfrac{1}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{4g} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2} = {v_0}^{2} \: {sin}^{2} \theta }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} }{2} = {v_0}^{2} }$

$\qquad$ $\longrightarrow \bf{v_0 = \sqrt{ \dfrac{ {v}^{2} }{2} } = \dfrac{v}{ \sqrt{2} } }$

Now, the vertical component of velocity of projectile at the height half of $\sf{h_m}$ will be –

$\qquad$ $\longrightarrow \bf{v_{(y)}=v_0 \: sin \theta }$

$\qquad$ $\longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} } \: sin \theta = \dfrac{v \: sin \: \theta}{ \sqrt{2} } }$

Therefore, the vertical component of velocity of projectile at this height will be–

☀️ $\qquad$ $\pink {\bf{ \dfrac{v \: sin \: \theta}{ \sqrt{2} }} }$

6 0

2 years ago

Read 2 more answers

If a 2 x 10^-4C test charge is given 6.5J of energy, determine the electric potential difference.

Gwar [14]

Answer:

The electric potential difference is 32500 volt.

Explanation:

Given that,

Charge $q=2\times10^{-4}C$

Energy = 6.5 J

We need to calculate the electric potential difference

Potential difference :

Potential difference is equal to the energy divide by charge.

Using formula of potential difference

$V=\dfrac{E}{Q}$

$V=\dfrac{6.5}{2\times10^{-4}}$

$V=32500\ volt$

Hence, The electric potential difference is 32500 volt.

8 0

2 years ago