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nevsk [136]
2 years ago
10

When a pair of 10-n forces act on a box of candy, the net force on the box is?

Physics
2 answers:
Grace [21]2 years ago
8 0
Any of the above depending on the direction of forces
Gnesinka [82]2 years ago
7 0
It can be anything from zero to 20N. It depends on the directions of the forces.
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Vectors A and B lie in the xy-plane. Vector A has a magnitude of 19.1 and is at an angle of 125.5º counterclockwise from the +x-
Nady [450]

Answer:

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Explanation:

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8 0
3 years ago
A container contains 200g of water at initial temperature of 30°C. An iron nail of mass 200g at temperature of 50°C is immersed
d1i1m1o1n [39]

Answer:

Assuming there is no heat loss to the surrounding.

Heat lost by iron equals heat gained by water.

0.2(450)(50-x)=0.2(4200)(x-30)

x=31.94 °C

Explanation:

4 0
3 years ago
Read 2 more answers
2.08
natka813 [3]

Calculate its average speed in meters per second

Answer:

5.77 m/s

Explanation:

Speed= Distance/Time

Distance= 40+ half of 40= 40+20= 60 m

Time= 8.8+1.6=10.4 s

Average speed= 60/10.4=5.769230769  m/s

Approximately, the average speed is 5.77 m/s

5 0
2 years ago
1. A car is slowing down at a constant rate from 40 m/s to come to a stop. The acceleration is - 5 m/s2. How
Oxana [17]

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Explanation:

5 0
2 years ago
Consider a father pushing a child on a playground merry-go-round. The system has a moment of inertia of 84.4 kg.m^2. The father
Sophie [7]

Answer:

Explanation:

Given that:

the initial angular velocity \omega_o = 0

angular acceleration \alpha = 4.44 rad/s²

Using the formula:

\omega = \omega_o+ \alpha t

Making t the subject of the formula:

t= \dfrac{\omega- \omega_o}{ \alpha }

where;

\omega = 1.53 \ rad/s^2

∴

t= \dfrac{1.53-0}{4.44 }

t = 0.345 s

b)

Using the formula:

\omega ^2 = \omega _o^2 + 2 \alpha \theta

here;

\theta = angular displacement

∴

\theta = \dfrac{\omega^2 - \omega_o^2}{2 \alpha }

\theta = \dfrac{(1.53)^2 -0^2}{2 (4.44) }

\theta =0.264 \ rad

Recall that:

2π rad = 1 revolution

Then;

0.264 rad = (x) revolution

x = \dfrac{0.264 \times 1}{2 \pi}

x = 0.042 revolutions

c)

Here; force = 270 N

radius = 1.20 m

The torque = F * r

\tau = 270 \times 1.20 \\ \\  \tau = 324 \ Nm

However;

From the moment of inertia;

Torque( \tau) = I \alpha \\ \\  Since( I \alpha) = 324 \ Nm. \\ \\  Then; \\ \\  \alpha= \dfrac{324}{I}

given that;

I = 84.4 kg.m²

\alpha= \dfrac{324}{84.4} \\ \\  \alpha=3.84 \ rad/s^2

For re-tardation; \alpha=-3.84 \ rad/s^2

Using the equation

t= \dfrac{\omega- \omega_o}{ \alpha }

t= \dfrac{0-1.53}{ -3.84 }

t= \dfrac{1.53}{ 3.84 }

t = 0.398s

The required time it takes= 0.398s

5 0
2 years ago
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