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steposvetlana [31]
3 years ago
8

A car moves with a speed of 72 km/h for 15 minutes and then with a speed of 80 km/h for the next 12 minutes. The total distance

covered by the car is:
Physics
2 answers:
sukhopar [10]3 years ago
8 0

Answer:

The total distance covered by the car is 3,810.08 m

Explanation:

Given;

initial speed of the car, u = 72 km/hr = 20 m/s

initial time, t₁ = 15 minutes = 900 s

final speed of the car, v = 80 km/hr = 22.22 m/s

final time, t₂ = 12 minutes = 720 s

The acceleration of the car is given as;

a = \frac{v-u}{t_2 -t_1} \\\\a = \frac{22.22-20}{900-720}\\\\a = 0.0123 \ m/s^2

The total distance covered by the car is given as;

v² = u² + 2as

where;

s is the total distance covered by the car

22.22² = 20² + 2(0.0123)s

22.22²  - 20² = 2(0.0123)s

93.728 = 0.0246s

s = 93.728 / 0.0246

s = 3,810.08 m

Therefore, the total distance covered by the car is 3,810.08 m

stiv31 [10]3 years ago
8 0

Answer:

34 km

Explanation:

Remark

You are dealing with distance not direction. So you just figure out the two parts separately and add.

15 minutes

t = 15 minutes * 1 hour / 60 minutes = 0.25 hours.

r = 72 km / hr

d = 72 km / hr * 0.25 hours

d = 18 km

12 minutes

t = 12 minutes * 1 hour/ 60 minutes = 0.2 hours

r = 80 km/hour

d = 80 * 0.2 = 16 km

Total distance

d = d1 + d2

d = 18 + 16 = 34 km

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uysha [10]

Answer:

c.a_m

Explanation:

We are given that

Acceleration due to gravity on the moon=a_m

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g_m=\frac{1}{6}g_e

Net force due to am on an object on moon=F_{net}=ma_m

There is no friction and no drag force and there is no gravity involved

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F=F_{net}(given)

ma_m=ma_e

a_e=a_m

Hence, option c is true.

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The cavity within a copper [β = 51 × 10-6 (C°)-1] sphere has a volume of 1.180 × 10-3 m3. Into this cavity is placed 1.100 × 10-
nikdorinn [45]

Answer:

The answer is "60.74^{\circ}".

Explanation:

Cavity and benzene should be extended in equal quantities.

\to 1.18 \times 10^{-3}\times (1+ \Delta T \times 0.000051) = 1.1\times 10^{-3} \times (1+ \Delta T \times 0.00124)\\\\\to  (\frac{1.18}{1.1})\times (1+ \Delta T \times 0.000051) = 1+ \Delta T \times 0.00124\\\\ \to 1.072\times (1+ \Delta T \times 0.000051) = 1+ \Delta T \times 0.00124\\\\ \to 1.072+ \Delta T \times 0.000054672 = 1+ \Delta T \times 0.00124\\\\ \to 1.072+ \Delta T \times 0.000054672 - 1- \Delta T \times 0.00124=0\\\\

\to 0.072+ \Delta T \times 0.000054672 - \Delta T \times 0.00124=0\\\\ \to 0.072+ \Delta T ( 0.000054672 -0.00124)=0\\\\ \to \Delta T ( 0.000054672 -0.00124)= -0.072\\\\ \to \Delta T = -\frac{0.072}{( 0.000054672 -0.00124)}\\\\ \to \Delta T = -\frac{0.072}{-0.001185328 }\\

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5 0
3 years ago
I NEED HELP ASAP!!!!!
Ratling [72]

Answer:

D) momentum of cannon + momentum of projectile= 0

Explanation:

The law of conservation of momentum states that the total momentum of an isolated system is constant.

In this case, the system cannon+projectile can be considered as isolated, because no external forces act on it (in fact, the surface is frictionless, so there is no friction acting on the cannon). Therefore, the total momentum of the two objects (cannon+projectile) must be equal before and after the firing:

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But the initial momentum is zero, because at the beginning both the cannon and the projectile are at rest:

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So the final momentum, which is sum of the momentum of the cannon and of the projectile, must also be zero:

p_f = p_{cannon}+p_{projectile} =0

6 0
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