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3 years ago

15

A car slows down from 65 km/s to 30 km/s in 5 seconds. What is its acceleration?

1 answer:

AveGali [126]3 years ago

7 0

Breaking is different from acceleration but he slows down at a speed of 7 kilometers per second.

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: A 70 kg man and a 12 kg sled are on the frictionless ice of a frozen lake, 25 m apart but connected by a rope of negligible ma

e-lub [12.9K]

Answer:

$x_1 = 3.74m$

Explanation:

given,

mass of man = 70 kg

mass of sled = 12 kg

F = m a_s

$a_s = \dfrac{F}{m}$

$a_s = \dfrac{8.2}{12}$

$a_s = 0.68\ m/s^2$

$F = m a_m$

$a_m = \dfrac{F}{m}$

$a_m = \dfrac{8.2}{70}$

$a_m = 0.12\ m/s^2$

$x_1+x_2 = 25$

$\dfrac{1}{2}a_ct^2+ \dfrac{1}{2}a_mt^2 = 25$

$(a_c+a_m)t^2=50$

$(0.12+0.68)t^2=50$

$t = \sqrt{\dfrac{50}{0.8}}$

t = 7.90 s

$x_1 = \dfrac{1}{2}a_ct^2$

$x_1 = 0.5\times 0.12 \times 7.90^2$

$x_1 = 3.74m$

5 0

3 years ago

A nasa spacecraft measures the rate r of at which atmospheric pressure on mars decreases with altitude. the result at a certain

Lesechka [4]

Answer: $4.21 \times 10^{-10} J/cm^4$

$1 kPa= 10^3 Pa$

$1 km=10^5 cm$

$1kPa/km=0.01 Pa/cm$

$1kPa/km=10^{-8} J/cm^4$

$\Rightarrow r= 0.0421 kPa/km= 0.0421 kPa/km \times \frac{10^{-8} J/cm^4}{1 kPa/km}= 0.0421 \times 10^{-8}J/cm^4=4.21 \times 10^{-10} J/cm^4$

3 0

3 years ago

Read 2 more answers

A car travels at a constant rate for 25 miles, going due east for one hour. Then it travels at a constant rate another 60 miles

egoroff_w [7]

60 mph east...........

6 0

2 years ago

Read 2 more answers

A 4.5-kg, three legged stool supports a 89-kg person. If each leg of the stool has a cross-sectional diameter of 2.8 cm and the

kkurt [141]

Answer:

4.96 × 10⁵ Pa

Explanation:

F = mg

$F = (m_{person}+m_{stool})g\\\\F = (4.5 + 89)*9.8\\\\F = 916.3 N$

This force is evenly distributed on the three leg

radius, r = d/2

= 2.8 / 2

= 1.4 cm = 0.014 m

total cross sectional area of the three legs, A = 3*pi*r^2

$= 3\times\pi \times0.014^2\\\\= 1.847\times10^-^3m^2$

Pressure due to weight,

P = Weight/A

$P = / 1.847 × 10⁻³\\P = \frac{ 916.3N}{1.847\times10^-^3} \\\\P= 496032.9Pa$

P = 4.96 × 10⁵ Pa

8 0

2 years ago

Of the following transitions in the Bohr hydrogen atom, the ________ transition results in the absorption of the highest-energy

8_murik_8 [283]

Answer:

The <em><u>n = 2 → n = 3</u></em> transition results in the absorption of the highest-energy photon.

Explanation:

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

Formula used for the radius of the $n^{th}$ orbit will be,

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number

Here: Z = 1 (hydrogen atom)

Energy of the first orbit in H atom .

$E_1=-13.6\times \frac{Z^2}{1^2} eV=-13.6 eV$

Energy of the second orbit in H atom .

$E_2=-13.6\times \frac{Z^2}{(2)^2} eV=-3.40 eV$

Energy of the third orbit in H atom .

$E_3=-13.6\times \frac{Z^2}{(9)^2} eV=-1.51 eV$

Energy of the fifth orbit in H atom .

$E_5=-13.6\times \frac{Z^2}{(2)^2} eV=-0.544 eV$

Energy of the sixth orbit in H atom .

$E_6=-13.6\times \frac{Z^2}{(2)^2} eV=-0.378 eV$

Energy of the seventh orbit in H atom .

$E_7=-13.6\times \frac{Z^2}{(2)^2} eV=-278 eV$

During an absorption of energy electron jumps from lower state to higher state.So, absorption will take place in :

1) n = 2 → n = 3

2) n= 5 → n = 6

Energy absorbed when: n = 2 → n = 3

$E=E_3-E_2$

$E=(-1.51 eV) -(-3.40 eV)=1.89 eV$

Energy absorbed when: n = 5 → n = 6

$E'=E_6-E_5$

$E'=(-0.378 eV)-(-0.544 eV) =0.166 eV$

1.89 eV > 0.166 eV

E> E'

So,the n = 2 → n = 3 transition results in the absorption of the highest-energy photon.

4 0

2 years ago