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Salsk061 [2.6K]

3 years ago

9

a bus travelling on a straight road at 25m/s accelerates uniformly at 5m/s squared for 2 seconds. find its speed in kilometres p

er hour

1 answer:

notka56 [123]3 years ago

5 0

Answer:

126 kmh⁻¹

Explanation:

We can simply solve this by applying motion equations

where

v - final velocity

u - initial velocity

a-acceleration

t - time

v = u + at

<em>= 25 + 5×2 = 35 ms⁻¹</em>

<em> = (35/1000)×3600 = 126 kmh⁻¹</em>

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A compact car has a maximum acceleration of 2.0 m/s2 when it carries only the driver and has a total mass of 1100 kg . you may w

nadya68 [22]

According to Newton`s law. Force exerted by car,

$F = m a = 1100 kg \times 2 m/s^2 = 2200 \ N$

After adding an additional 400 kg of mass, the force will be same therefore the acceleration

$F = 2200 \ N = (1100 \ kg + 400 \ kg) a \\\\ a = \frac{2200 \ N}{1500 \ kg} = 1.47 \ m/s^2$

Thus, the acceleration after adding the masses is 1.47 \ m/s^2.

4 0

3 years ago

Determine the velocity required for a moving object 2.00 x 10^4 m above the surface of Mars to escape from Mars's gravity. The m

Ket [755]

Answer:

Therefore the escape velocity from Mar's gravity is $15.88 \times 10^4$ m/s.

Explanation:

Escape velocity: Escape velocity is a the minimum velocity that a object needs to escape from the gravitational field of massive body.

$V_{escape}=\sqrt{\frac{2GM}{R}}$

$V_{escape}=$ Escape velocity

G=Universal gravitational constant = 6.673×10⁻¹¹N m²/Kg²

M= mass of Mars = 6.42×10²³ kg

R = Radius of the Mars = 3.40×10³m

The escape velocity does not depend on the velocity of a object.

$V_{escape}=\sqrt{\frac{2\times6.673\times 10^{-11}\times 6.42\times 10^{23}}{3.40\times10^3}}$

$=15.88 \times 10^4$ m/s

Therefore the escape velocity from Mar's gravity is $15.88 \times 10^4$ m/s.

3 0

3 years ago

What happens when light enters water?

Svet_ta [14]

The answer should be A

8 0

3 years ago

Can any ideal gas power cycle have a thermal efficiency greater than 55 percent when using thermal energy reservoirs at 627∘C at

Yanka [14]

Answer:

Since the maximum thermal efficiency is higher than 55 percent, there can be a power cycle with these reservoir temperature with an efficiency higher than 55 percent.

Explanation:

The maximum thermal efficiency is determined from the given temperature

nth Carnot = 1- TL/TH

Where TL= 17+273= 290k

TH= 627*273= 900K.

nth Carnot = 1- 290/900 = 0.68

0.68*100 = 68 percent

8 0

3 years ago

a particle with a charge of 5.5 x 10^-8 c is 3.5 cm from a particle with a charge of -2.3 x10^-8 c. the potential energy of this

Yuri [45]

Answer:

-32.5 * 10^-5 J

Explanation:

The potential energy of this system of charges is;

Ue = kq1q2/r

Where;

k is the Coulumb's constant

q1 and q2 are the magnitudes of the charges

r is the distance of separation between the charges

Substituting values;

Ue = 9.0×10^9 N⋅m2/C2 * 5.5 x 10^-8 C *( -2.3 x10^-8) C/(3.5 * 10^-2)

Ue= -32.5 * 10^-5 J

4 0

3 years ago