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2 years ago

Can someone please help me

Physics

1 answer:

pickupchik [31]2 years ago

7 0

Answer:C (198 seconds)

Explanation: The cyclist makes the first lap in (180.00 - 6.00) = 174.00 seconds. The average time per lap for all three circuits is (600.00 - 6.00) = 594/3 = 198 seconds.

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Magnetic field 0.02 mfrom a wire is 0.1 T. What is the magnitude of the magnetic field of 0.01 m from the same wire?

Gre4nikov [31]

Answer:

magnetic field from the same wire

6 0

2 years ago

Groups of students are making measurements of small cars rolling across a table. the class has measuring tapes with markings in

TiliK225 [7]

To find the accurate measurement of small cars, the teacher asks students to make all the measurements in centimeters.

Centimeters Measurements:

A centimeter is a metric unit of measurement used for measuring the length of an object, It is written as cm
Centimeter is one hundredth of a meter, 1 meter is 0.01 cm.

Inches measurements:

An inch can be defined as a unit of length in the customary system of measurement. Length in inches is either represented by in or ''.
1 meter is equal to 39.37 inches

here, the cars are small objects.

The number of centimeters is always bigger,

because a centimeter unit is smaller than an inch unit, and it takes more of them when we are measuring.

Hence,

To find the accurate measurement of small cars, the teacher asks students to make all the measurements in centimeters.

Learn more about accurate measurement here:

brainly.com/question/4119127

#SPJ4

4 0

2 years ago

A proton moves through a magnetic field at 20.3 % of the speed of light. At a location where the field has a magnitude of 0.0062

galina1969 [7]

Answer:

Force on the proton will be $.73\times 10^{-14}N$

Explanation:

We have given speed of proton is 20.3% of speed of light

Speed of light $c=3\times 10^8m/sec$

So speed of proton $v=\frac{3\times 10^8\times 23}{100}=6.9\times 10^7m/sec$

Magnetic field B = 0.00629 T

Charge on proton $q=1.6\times 10^{-16}C$

Angle between velocity and magnetic field $\Theta =137^{\circ}$

Force on the proton is equal to $F=qvBsin\Theta =1.6\times 10^{-19}\times 6.9\times 10^7\times 0.00629\times sin(137^{\circ})=4.73\times 10^{-14}N$

4 0

3 years ago

A block with mass m =6.9 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.24 m.

Burka [1]

U need to set up n solve the general eqn for simple harmonic motion:
x" = -(k/m)x

solution is x(t) = (x0)*cos(wt) + (v0/w)*sin(wt)
where w=sqrt(k/m), x0 is x-position at t=0 and v0 is vel at t=0
u already calculated f in Q.2 and w = 2*pi*f
x0 is 0 as it starts at eqm
v0 is given at 5.1

so u have x(t)

vel is given by x'(t) = (x0)*(-w)*sin(wt) + (v0/w)*w*cos(wt)

substitute t=0.32, x0=0, v0=5.1 n w in the above, u can solve for v at t=0.32.

5 0

3 years ago

For what value m of the clockwise couple will the horizontal component ax of the pin reaction at a be zero? if a couple of that

sergejj [24]

Thank you for posting your question here at brainly. Below is the answer:

sum of Mc = 0 = -Ay(4.2 + 3cos(59)) + (275)(2.1 + 3cos(59)) + M
- Ay = (M + (275*(2.1 + 3cos(59)))/(4.2 + 3cos(59)) 

sum of Ma = 0 = (-275)(2.1) - Cy(4.2 + 3cos(59)) + M 
- Cy = (M - (275*2.1))/(4.2 + 3cos(59)) 

Ay + Cy = 275 = ((M+1002.41)+(M-577.5))/(5.745) 
= (2M + 424.91)/(5.745) 

M = ((275*5.745) - 424.91)/2 
= 577.483 which rounds off to 577 

Is it maybe supposed to be Ay - Cy = 275

8 0

3 years ago

Can someone please help me​

Can someone please help me