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Nataly [62]
4 years ago
5

A car, moving along a straight stretch of highway, begins to accelerate at 0.0323 m/s 2 . It takes the car 63.3 s to cover 1 km.

How fast was the car going when it first began to accelerate? Answer in units of m/s.
Physics
1 answer:
rosijanka [135]4 years ago
7 0
X=ut+0.5at²
1000 = 63.3u + 0.5 * 0.0323 * 63.3²
u = (1000 - 0.1615*63.3²)/63.3
u = 5.57 m/s

i don't even know if this is correct
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A famous physics tale is about a rich man who was found dead. He was
anyanavicka [17]
<h2>Answer: C) He could have thrown the bag of money sideways, creating a  horizontal reaction force on himself.</h2>

Explanation:

According to Newton's third law of motion, when two bodies interact between them, appear equal forces and opposite senses in each of them.

To understand it better:

Each time a body or object exerts a force on a second body or object, it (the second body) will exert a force of equal magnitude and direction but in the opposite direction on the first.

So, if the rich man had pushed the bag of money horizontally opposite of where he was, he could have saved himself.

7 0
3 years ago
A spring hangs from the ceiling with an unstretched length of x 0 = 0.69 m x0=0.69 m . A m 1 = 7.5 kg m1=7.5 kg block is hung fr
sergejj [24]

Answer:

x2=0.732m

Explanation:

We can calculate the spring constant using the equilibrium equation of the block m1. Since the spring is in equilibrium, we can say that the acceleration of the block is equal to zero. So, its equilibrium equation is:

m_1g-k\Delta x_1=0\\\\\implies k=\frac{m_1g}{\Delta x_1}\\\\k=\frac{(7.5kg)(9.8m/s^{2})}{0.84m-0.69m}=490N/m

Then using the equilibrium equation of the block m2, we have:

m_2g-k\Delta x_2=0\\\\\\implies x_2=x_0+\frac{m_2g}{k} \\x_2=0.69m+\frac{(2.1kg)(9.8m/s^{2})}{490N/m}= 0.732m

In words, the lenght x2 of the spring when the m2 block is hung from it, is 0.732m.

6 0
3 years ago
Convert 27,549 into scientific notation
dezoksy [38]
2.7549 x 10^4 is the answer I hope this helped u
7 0
3 years ago
On a cello, the string with the largest linear density (1.44 x 10-2 kg/m) is the C string. This string produces a fundamental fr
Rus_ich [418]

Answer:

T=346.5N

Explanation:

From the question we are told that:

Density \rho=  (1.44 * 10^{-2} kg/m)

FrequencyF=93.0Hz

Lengthl=0.834m

Generally the equation for Frequency is mathematically given by

F=\frac{1}{2 l}sqrt{\frac{T}{\rho}}

Therefore

T=\rho(2 lF)^2

T= (1.44 * 10^{-2}*(2*0.834)(93.0))^2

T=346.5N

4 0
3 years ago
A river flows due east at 1.60 m/s. A boat crosses the river from the south shore to the north shore by maintaining a constant v
Citrus2011 [14]

Answer:

part (a) v\ =\ 10.42\ at\ 81.17^o towards north east direction.

part (b) s = 46.60 m

Explanation:

Given,

  • velocity of the river due to east = v_r\ =\ 1.60\ m/s.
  • velocity of the boat due to the north = v_b\ =\ 10.3\ m/s.

part (a)

River is flowing due to east and the boat is moving in the north, therefore both the velocities are perpendicular to each other and,

Hence the resultant velocity i,e, the velocity of the boat relative to the shore is in the North east direction. velocities are the vector quantities, Hence the resultant velocity is the vector addition of these two velocities and the angle between both the velocities are 90^o

Let 'v' be the velocity of the boat relative to the shore.

\therefore v\ =\ \sqrt{v_r^2\ +\ v_b^2}\\\Rightarrow v\ =\ \sqrt{1.60^2\ +\ 10.3^2}\\\Rightarrow v\ =\ 10.42\ m/s.

Let \theta be the angle of the velocity of the boat relative to the shore with the horizontal axis.

Direction of the velocity of the boat relative to the shore.\therefore Tan\theta\ =\ \dfrac{v_b}{v_r}\\\Rightarrow Tan\theta\ =\ \dfrac{10.3}{1.60}\\\Rightarrow \theta\ =\ Tan^{-1}\left (\dfrac{10.3}{1.60}\ \right )\\\Rightarrow \theta\ =\ 81.17^o

part (b)

  • Width of the shore = w = 300m

total distance traveled in the north direction by the boat is equal to the product of the velocity of the boat in north direction and total time taken

Let 't' be the total time taken by the boat to cross the width of the river.\therefore w\ =\ v_bt\\\Rightarrow t\ =\ \dfrac{w}{v_b}\\\Rightarrow t\ =\ \dfrac{300}{10.3}\\\Rightarrow t\ =\ 29.12 s

Therefore the total distance traveled in the direction of downstream by the boat is equal to the product of the total time taken and the velocity of the river\therefore s\ =\ u_rt\\\Rightarrow s\ =\ 1.60\times 29.12\\\Rightarrow s\ =\ 46.60\ m

7 0
3 years ago
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