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2 years ago

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:

a!aaaaaaaaaaaa!aaaaaa

Explanation:

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]

Answer:

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

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

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