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Alexandra [31]
3 years ago
15

A crate of oranges with a total mass of 6.7 kg is being pulled across a skating rink (frictionless) with a rope that makes an an

gle of 0.7 rad above the horizontal. John, who measures the motion of the crate, sees that it accelerates at a constant rate of 4.2 m/s2. What is the tension in the rope in N?
Physics
2 answers:
AVprozaik [17]3 years ago
5 0

Answer:

The tension in the rope is 36.792 N

Explanation:

Data given:

m = 6.7 kg

θ = 0.7 rad = 40.107°

a = acceleration = 4.2 m/s²

The tension in the rope is:

T=\frac{ma}{cos\theta } =\frac{6.7*4.2}{cos40.107} =36.792N

Mama L [17]3 years ago
4 0

Answer:

36.74 N

Explanation:

Given that:

A crate of oranges with a total mass  (m) = 6.7 kg

angle θ = 0.7 rad

angle θ = 0.7 * \frac{180}{\pi}

angle θ = 40°

acceleration = 4.2 m/s²

Given that:

T cosθ  = ma

T cos 40° = 6.7 × 4.2

T = \frac { 6.7 * 4.2}{cos \ 40}

T = \frac { 28.14}{0.7660}

T = 36.74 N

Thus, the tension in the rope = 36.74 N

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lisov135 [29]

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Hope this helped!

7 0
3 years ago
Give an example of a compound machine. Explain how at least two simple machines are part of this complex machine.
11Alexandr11 [23.1K]

Answer:

Bicycle

Explanation:

A compound machine is a machine which is a combination of simple machines.

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6 0
3 years ago
Four particles are in a 2-d plane with masses, x- and y- positions, and x- and y- velocities as given in the table below: what i
Arte-miy333 [17]
I attached the picture of the missing table.
Center of mass is the point such that if you apply force to it, the system would move without rotating.
We can use following formula to calculate the center of mass:
R=\frac{1}{M}\sum_{i=1}^{n=i}m_ir_i
Where M is the sum of the masses of all particles.
Part 1
To calculate the x coordinate of the center of mass we will use this formula:
R_x=\frac{1}{M}\sum_{i=1}^{n=i}m_ix_i
I will do all the calculations in the google sheet that I will share with you.
For the x coordinate of the center of mass we get:
R_x=0.96m
Part 2
To calculate the y coordinate of the center of mass we will use this formula:
R_y=\frac{1}{M}\sum_{i=1}^{n=i}m_iy_i
I will do all the calculations in the google sheet that I will share with you.
For the x coordinate of the center of mass we get:
R_y=-0.84m
Part 3
We will calculate speed along x and y-axis separately and then will add them together.
v_x=\frac{\sum_{i=1}^{n=i}m_iv_x_i}{M}
v_y=\frac{\sum_{i=1}^{n=i}m_iv_y_i}{M}
Total velocity is:
v=\sqrt{v_x^2+v_y^2}
Once we calculate velocities we get:
v_x=-1.08\frac{m}{s}\\ v_y=-0.03\frac{m}{s}\\ v=\sqrt{(-1.08)^2+(-0.03)^2}=1.08\frac{m}{s}
Part 4
Because origin is left to our center of mass(please see the attached picture) placing fifth mass in the origin would move the center of mass to the left along the x-axis.
Part 5
If you place fifth mass in the center of the mass nothing would change. The center of mass would stay in the same place.
Here is the link to the spreadsheet:
https://docs.google.com/spreadsheets/d/1SkQHbI1BxiJnwpWbLmP0XWgcNPrGquH1K2MfN6cznVo/edit?usp=sharing

3 0
3 years ago
Please help me to do this problem
Novosadov [1.4K]

Answer:

we got time and velocity over time.

so the distance is again the area underneath the graph

for a triangle with known base and height it's

4*10 / 2

distance traveled is 20

deceleration occurs when velocity decreases. that happens from t=2 till t=4

in 2 time-units we loose 10 units of velocity, so we decelerate by 5 units per 1 time

a (from t=2 to t=4) = -5v/t

7 0
3 years ago
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