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alexdok [17]
3 years ago
10

A runner runs 2.0 km in 5.0 minutes

Physics
2 answers:
nata0808 [166]3 years ago
6 0
But what do we need to do??
Nadya [2.5K]3 years ago
4 0

Answer: What is this supposed to be converted into?

Explanation:

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Calculate acceleration of an object moving from a stop to 21 m/s in 3.9 s
Alex
5.38
21 divided by 3.9 is 5.38 and the equation for acceleration is change in velocity divided by time
7 0
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What happens to light when it strikes the inside surface of a smooth, curved mirror?
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<span>It bounces back toward a single spot.  This is how the focusing mirrors on headlights work</span>
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Can someone please help me with these physics problems? I just don’t even know where to start.
KIM [24]

#1

for the block of mass 5 kg normal force is given as

F_n = mg

F_n = 5*9.8 = 49 N

friction force is given as

F_f = \mu F_n

F_f = 0.1*49 = 4.9 N

Net force is given as

F_{net} = ma

F_{net} = 5*2 = 10 N

now we know that

F_{net} = F_{app} - F_f

10 = F_{app} - 4.9

F_{app} = 14.9 N

#2

Normal force is given as

F_n = mg

F_n = 6*9.8

F_n = 58.8 N

now we know that

F_{net} = F_{app} - F_f

F_{net} = 0

as object moves with constant velocity

F_{app} = F_f = 15 N

now for coefficient of friction we can use

F_f = \mu F_n

15 = \mu * 58.8

\mu = 0.255

#3

net force upwards is given as

F = 1.2 * 10^{-4} N

mass is given as

m = 7 * 10^{-5} kg

now as per newton's law we can say

F = ma

1.2 * 10^{-4} = 7 * 10^{-5} * a

a = 1.71 m/s^2

#4

As we know that when block is sliding on rough surface

part a)

net force = applied force - frictional force

F_{net} = F_{app} - F_f

ma = F_{app} - F_f

5*6 = 40 - F_{f}

F_f = 40 - 30 = 10 N

part b)

for coefficient of friction we can use

F_f = \mu F_n

10 = \mu * F_n

here normal force is given as

F_n = mg = 5*9.8 = 49 N

now we have

\mu = \frac{10}{49} = 0.204

#5

if an object is initially at rest and moves 20 m in 5 s

so we can use kinematics to find out the acceleration

d = v_i*t + \frac{1}{2}at^2

20 = 0 + \frac{1}{2}a(5^2)

a = 1.6 m/s^2

now net force is given as

F_{net} = ma

F_{net} = 10*1.6 = 16 N

#6

an object travelling with speed 25 m/s comes to stop in 1.5 s

so here acceleration of object is given as

a = \frac{v_f - v_i}{t}

a = \frac{0 - 25}{1.5} = -16.67 m/s^2

now the force is gievn as

F = ma

F = 5*16.67 = 83.3 N

3 0
4 years ago
A circular loop of flexible iron wire has an initial circumference of 164cm , but its circumference is decreasing at a constant
Travka [436]

Answer:

emf = 0.02525 V

induced current with a counterclockwise direction

Explanation:

The emf is given by the following formula:

emf=-\frac{\Delta \Phi_B}{\Delta t}=-B\frac{\Delta A}{\Delta t}\ \ =-B\frac{A_2-A_1}{t_2-t_1}   (1)

ФB: magnetic flux =  BA

B: magnitude of the magnetic field = 1.00T

A2: final area of the loop; A1: initial area

t2: final time, t1: initial time

You first calculate the final A2, by taking into account that the circumference of loop decreases at 11.0cm/s.

In t = 4 s the final circumference will be:

c_2=c_1-(11.0cm/s)t=164cm-(11.0cm/s)(4s)=120cm

To find the areas A1 and A2 you calculate the radius:

r_1=\frac{164cm}{2\pi}=26.101cm\\\\r_2=\frac{120cm}{2\pi}=19.098cm

r1 = 0.261 m

r2 = 0.190 m

Then, the areas A1 and A2 are:

A_1=\pi r_1^2=\pi (0.261m)^2=0.214m^2\\\\A_2=\pi r_2^2=\pi (0.190m)^2=0.113m^2

Finally, the emf induced, by using the equation (1), is:

emf=-(1.00T)\frac{(0.113m^2)-(0.214m^2)}{4s-0s}=0.0252V=25.25mV

The induced current has counterclockwise direction, because the induced magneitc field generated by the induced current must be opposite to the constant magnetic field B.

4 0
3 years ago
A cyclist is travelling at 4 metres per second for 100 metres,<br> how long does it take?
Slav-nsk [51]

Answer:

25 seconds!

Explanation:

I just had to find what number to multiply 4 to get 100 and it was 25

4x25=100

Or you could do the easier way by dividing 100 by 4 which has the same answer: 25

100÷4=25

7 0
4 years ago
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