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kolbaska11 [484]

3 years ago

5

A factory conveyor belt rolls at 3 m/s. A mouse sees a piece of cheese directly across the belt and heads straight for the chees

e at 4 m/s. What is the mouse's speed relative to the factory floor?

1 answer:

olasank [31]3 years ago

7 0

Answer:

The speed of the mouse relative to the factory floor is 5 m/s.

Explanation:

Given that,

A factory conveyor belt rolls at 3 m/s, $v_1=3\ m/s$

A mouse sees a piece of cheese directly across the belt and heads straight for the cheese at 4 m/s, $v_2=4\ m/s$

To find,

The mouse's speed relative to the factory floor.

Solution,

The resultant velocity of the mouse is given by the resultant of two velocities. It is given by :

$v=\sqrt{v_1^2+v_2^2}$

$v=\sqrt{3^2+4^2}$

v = 5 m/s

So, the speed of the mouse relative to the factory floor is 5 m/s.

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How fast can the 140 a current through a 0.200 h inductor be shut off if the induced emf cannot exceed 80.0 v?

Vesna [10]

Recall that to compute for the emf of a circuit given current and inductance, we must recall that

$emf = - M \frac{\Delta I }{\Delta t}$

where I is the current (A), M is the mutual inductance (h), and t is the time (ms). Since the current must not exceed 80.0 V, we have

$80.0 \geq 0.200(\frac{140}{t})$
$t \geq \frac{28.0}{80}$
$t \geq 0.35$

From this, we see that it must take at least 0.35 ms so it doesn't exceed 80 V.
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2 years ago

To test the performance of its tires, a car

velikii [3]

<u>Answer</u>:

The coefficient of static friction between the tires and the road is 1.987

<u>Explanation</u>:

<u>Given</u>:

Radius of the track, r = 516 m

Tangential Acceleration $a_r$ = 3.89 m/s^2

Speed,v = 32.8 m/s

<u>To Find:</u>

The coefficient of static friction between the tires and the road = ?

<u>Solution</u>:

The radial Acceleration is given by,

$a_{R = \frac{v^2}{r}$

$a_{R = \frac{(32.8)^2}{516}$

$a_{R = \frac{(1075.84)}{516}$

$a_{R = 2.085 m/s^2$

Now the total acceleration is

$\text{ total acceleration} = \sqrt{\text{(tangential acceleration)}^2 +{\text{(Radial acceleration)}^2$

=> $= \sqrt{ (a_r)^2+(a_R)^2}$

=> $\sqrt{ (3.89 )^2+( 2.085)^2}$

=> $\sqrt{ (15.1321)+(4.347)^2}$

=> $19.4791 m/s^2$

The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of static friction is

$\mu =\frac{f}{W}$

From (1) and (2)

$\mu =\frac{ma}{mg}$

$\mu =\frac{a}{g}$

Substituting the values, we get

$\mu =\frac{19.4791}{9.8}$

$\mu =1.987$

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3 years ago

What magnitude charge creates a 1.0 n/c electric field at a point 1.0 m away?

Stolb23 [73]

Answer:

$1.1\cdot 10^{-10}C$

Explanation:

The electric field produced by a single point charge is given by:

$E=k\frac{q}{r^2}$

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have

E = 1.0 N/C (magnitude of the electric field)

r = 1.0 m (distance from the charge)

Solving the equation for q, we find the charge:

$q=\frac{Er^2}{k}=\frac{(1.0 N/c)(1.0 m)^2}{9\cdot 10^9 Nm^2c^{-2}}=1.1\cdot 10^{-10}C$

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2 years ago

A proton traveling due west in a region that contains only a magnetic field experiences a vertically upward force (away from the

Zanzabum

Answer:

D

Explanation:

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2 years ago

Read 2 more answers

A 0.111 kg hockey puck moving at 55 m/s is caught by a 80. kg goalie at rest. with what speed does the goalie slide on the frict

madreJ [45]

Answer:

0.076 m/s

Explanation:

Momentum is conserved:

m v = (m + M) V

(0.111 kg) (55 m/s) = (0.111 kg + 80. kg) V

V = 0.076 m/s

After catching the puck, the goalie slides at 0.076 m/s.

8 0

3 years ago