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2 years ago

A car accelerates at a rate of 8.8 m/s² with a force from the tires of 15,840 N.

Physics

1 answer:

dezoksy [38]2 years ago

4 0

Answer:

1,800kg

Explanation:

Force = Mass x Acceleration

F = m x a

15840N = m x 8.8

8.8 x m = 15840

m = 15840/8.8

= 1,800kg

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A proton is moving at 425 m/s. (a) How much work must be done on it to stop it? (A proton has a mass of 1.67×10−27 kg.) (b) Assu

Tpy6a [65]

Answer:

a)1.51*10^-22joules b) 1.89*10^-7m

Explanation:

Work done to stop the proton = the kinetic energy of the proton = 1/2 mv^2 = 1/2* 1.67*10^-27* 425*425 = 1.51* 10 ^ -22 joules

b) net force acting to stop the proton = 8.01*10^-16

Work done needed to stop the proton = net force acting opposite the motion * distance

Distance covered = need work done/ net force

Distance = 1.51*10^-22/8.01*10^-16= 1.89*10^-7m

8 0

3 years ago

Suppose a cup of coï¬ee is at 100 degrees Celsius at time t = 0, it is at 70 degrees at t = 10 minutes, and it is at 50 degrees

Alexandra [31]

Answer:

T ambient = 10 degrees

Explanation:

Using Newton's Law of Cooling:

T(t) = Tamb + (Ti - Tamb)*e^(-kt) ..... Eq 1

Ti = 100

We have two points to evaluate the above equation as follows:

T = 70 @ t = 10 using Eq 1

70 = Tamb + (100 - Tamb)*e^(-10k) ... Eq 2

T = 50 @ t = 20 using Eq 1

50 = Tamb + (100 - Tamb)*e^(-20k) ... Eq 3

Solving the above Eq 2 and Eq 3 simultaneously:

Using Eq 2:

(70 - Tamb) / (100 - Tamb) = e^(-10k)

Squaring both sides we get:

((70 - Tamb) / (100 - Tamb))^2 = e^(-20k) .... Eq 4

Substitute Eq 4 into Eq 3

50 = Tamb + (100 - Tamb)*((70 - Tamb) / (100 - Tamb))^2

After simplification:

50 = (Tamb (100-Tamb) + (70-Tamb)^2) / (100 - Tamb)

5000 - 50*Tamb = 4900 - 40*Tamb

Tamb = 100 / 10 = 10 degrees

6 0

3 years ago

Find the net charge of a system consisting of 6.25×10^6 electrons and 7.75×10^6 protons. Express your answer using three signifi

jok3333 [9.3K]

Answer:

2.40 x 10⁻¹³ C

Explanation:

$n_{e}$ = number of electrons = 6.25 x 10⁶

$q_{e}$ = charge on electron = - 1.6 x 10⁻¹⁹ C

$n_{p}$ = number of protons = 7.75 x 10⁶

$q_{p}$ = charge on proton = 1.6 x 10⁻¹⁹ C

Net charge is given as

Q = $n_{e}$ $q_{e}$ + $n_{p}$ $q_{p}$

Q = (- 1.6 x 10⁻¹⁹) (6.25 x 10⁶) + (1.6 x 10⁻¹⁹) (7.75 x 10⁶)

Q = 2.40 x 10⁻¹³ C

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3 years ago

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Explanation:

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3 years ago

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alexandr1967 [171]

Explanation:

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3 years ago