What does the area under a speed-time graph represent

Need help on physics/momentum

koban [17]

Answer:

vf=11.67m/s

Explanation:

vf:

Δp=m(vf-vi)

80=12(vf-5)

80=12vf-60

140=12vf

vf=11.67m/s

pf:

p=m*v

p=12*11.67

p=140.04 N*s

Hope this helps

5 0

3 years ago

Using the periodic table, predict the formulas of stable ionic compounds. Select THREE that are correct.

AleksandrR [38]

The first one,second one and the forth one are correct

6 0

3 years ago

Read 2 more answers

A total electric charge of 6.75 nC is distributed uniformly over the surface of a metal sphere with radius 20.0 cm. If the poten

djverab [1.8K]

Answer:

a) 60 V

b) 125 V

c) 125 V

Explanation:

Given

We are given the total electric charge q = 6.75 nC = 6.75x 10^-9 C distributed uniformly over the surface of a metal sphere with a radius of R = 20.0 cm = 0.020 m.

Required

We are asked to calculate the potential at the distances

(a) r = 10.0 cm

(b) r = 20.0 cm

(c) r = 40.0 cm

Solution

(a) Here, the distance r > R so, we can get the potential outside the sphere (r > R) where the potential is given by

V = q/4 $\pi$ ∈_o (1)

r is the distance where the potential is measured and the term 1/4 $\pi$ ∈_o equals 9.0 x 10^9 Nm^2/C^2. Now we can plug our values for q and r into equation (1) to get the potential V where r = 0.10 m

V= 1*q/4 $\pi$ ∈_o*r

=60 V

(b) Here the distance r is the same for the radius R, so we can get the potential inside the sphere (r = R) where the potential is given by

V = 1*q/4 $\pi$ ∈_o*R (2)

Now we can plug our values for q and R into equation (2) to get the potential V where R = 0.20 m

V = 1*q/4 $\pi$ ∈_o*R

= 125 V

(c) Inside the sphere the electric field is zero therefore, no work is done on a test charge that moves from any point to any other point inside the sphere. Thus the potential is the same at every point inside the sphere and is equal to the potential on the surface. and it will be the same as in part (b)

V= 125 V

7 0

3 years ago

PLEASE SOLVE QUICKLY!!! Solve for A, B, and C from graph

hram777 [196]

A = 59.35cm

B = 196.56g

C = 74.65g

Explanation:

We know,

$x = \frac{L}{\frac{W}{F} +1}$

and L = x+y

1.

Total length, L = 100cm

Weight of Beam, W = 71.8g

Center of mass, x = 49.2cm

Added weight, F = 240g

Position weight placed from fulcrum, y = ?

$L-y = \frac{L}{\frac{W}{F}+1 } \\100 - y = \frac{100}{\frac{71.8}{49.2}+1 } \\100 - y = \frac{100}{1.46+1}\\\\100 - y = \frac{100}{2.46} \\100-y = 40.65\\\\y = 59.35cm$

Therefore, position weight placed from fulcrum is 59.35cm

2.

Total length, L = 100cm

Center of mass, x = 47.8 cm

Added weight, F = 180g

Position weight placed from fulcrum, y = 12.4cm

Weight of Beam, W = ?

$47.8 = \frac{100}{\frac{W}{180}+1 }\\\47.8 = \frac{100}{\frac{W+180}{180} } \\\\47.8 = \frac{100 X 180}{W+180}\\ \\47.8W + 47.8 X 180 = 18000\\47.8W = 18000 - 8604\\W = \frac{9396}{47.8}\\ W = 196.56g$

Therefore, weight of the beam is 196.56g

3.

Total length, L = 100cm

Center of mass, x = 50.8 cm

Position weight placed from fulcrum, y = 9.8cm

Weight of Beam, W = 72.3g

Added weight, F = ?

$50.8 = \frac{100}{\frac{72.3}{F}+1 }\\\ 50.8 = \frac{100}{\frac{72.3+F}{F} } \\\\50.8 = \frac{100 X F}{72.3+F}\\ \\50.8 X 72.3 + 50.8 X F = 100F\\\\3672.84 = 100F-50.8F\\3672.84 = 49.2F\\F = 74.65g$

Therefore, Added weight F is 74.65g

A = 59.35cm

B = 196.56g

C = 74.65g

4 0

3 years ago

Chemistry determine the pressure

sashaice [31]

I think it's 1.03412969 or 1.03

8 0

4 years ago

What does the area under a speed-time graph represent​

What does the area under a speed-time graph represent