Ask question

Ask question

All categories

IrinaVladis [17]

3 years ago

14

During a football game, the running back changes direction from running 4.5 m/s [N] to 4.5 m/s [S] over 8 s. What is the acceler

ation of the running back during that time?

1 answer:

Nimfa-mama [501]3 years ago

8 0

Explanation:

Take north to be positive and south to be negative.

a = (v − v₀) / t

a = (-4.5 m/s − 4.5 m/s) / 8 s

a = -1.125 m/s²

The acceleration is 1.125 m/s² south.

You might be interested in

Can u plz help I'm failing and I have 2 weeks to get my grades up someone help me plz plz plz plz this is not a test ok it was h

Anastaziya [24]

1. B
2. A
3. B
4. C
5. heterogeneous
6. element
7. do
8. true
9. true
10. true

7 0

3 years ago

Read 2 more answers

For a magnetic field strength of 2T, estimate the magnitude of the maximum force on a 1-mm-long segment of a single cylindrical

nydimaria [60]

Answer:

Incomplete question

This is the complete question

For a magnetic field strength of 2 T, estimate the magnitude of the maximum force on a 1-mm-long segment of a single cylindrical nerve that has a diameter of 1.5 mm. Assume that the entire nerve carries a current due to an applied voltage of 100 mV (that of a typical action potential). The resistivity of the nerve is 0.6ohms meter

Explanation:

Given the magnetic field

B=2T

Lenght of rod is 1mm

L=1/1000=0.001m

Diameter of rod=1.5mm

d=1.5/1000=0.0015m

Radius is given as

r=d/2=0.0015/2

r=0.00075m

Area of the circle is πr²

A=π×0.00075²

A=1.77×10^-6m²

Given that the voltage applied is 100mV

V=0.1V

Given that resistive is 0.6 Ωm

We can calculate the resistance of the cylinder by using

R= ρl/A

R=0.6×0.001/1.77×10^-6

R=339.4Ω

Then the current can be calculated, using ohms law

V=iR

i=V/R

i=0.1/339.4

i=2.95×10^-4 A

i=29.5 mA

The force in a magnetic field of a wire is given as

B=μoI/2πR

Where

μo is a constant and its value is

μo=4π×10^-7 Tm/A

Then,

B=4π×10^-7×2.95×10^-4/(2π×0.00075)

B=8.43×10^-8 T

Then, the force is given as

F=iLB

Since B=2T

F=iL(2B)

F=2.95×10^-4×2×8.34×10^-8

F=4.97×10^-11N

7 0

3 years ago

You pull a solid iron ball with a density of 7.86 g/cm3 and a radius of 1.50 cm upward through a fluid at a constant speed of 9.

erastova [34]

Answer:

F = 1.099N

Explanation:

See the attachment below.

6 0

3 years ago

Which of these statements is true?

9966 [12]

C.) Meiosis involves two cycles of cell division

Hope this helps!

6 0

3 years ago

Read 2 more answers

The three forces shown act on a particle. what is the direction of the resultant of these three forces?

melisa1 [442]

Missing figure: http://d2vlcm61l7u1fs.cloudfront.net/media/f5d/f5d9d0bc-e05f-4cd8-9277-da7cdda3aebf/phpJK1JgJ.png

Solution:
We need to find the magnitude of the resultant on both x- and y-axis.

x-axis) The resultant on the x-axis is
$F_x = 65 N\cdot cos 30^{\circ} - 30 N - 20 N\cdot sin 20^{\circ} = 19.45 N$
in the positive direction.

y-axis) The resultant on the y-axis is
$F_y = 65 N \cdot sin 30^{\circ} - 20 N \cdot cos 20^{\circ} = 13.70 N$
in the positive direction.

Both Fx and Fy are positive, so the resultant is in the first quadrant. We can find the angle and so the direction using
$\tan \alpha = \frac{F_y}{F_x} = \frac{13.70 N}{19.45 N}=0.7$
from which we find
$\alpha=35^{\circ}$

7 0

3 years ago