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irina [24]
2 years ago
11

a car accelerates along a straight road from rest to 75km/h in 5.0s. What is the magnitude of its average acceleration? show up

solutions​
Physics
1 answer:
kykrilka [37]2 years ago
5 0

Answer:

4.186 m/s^2

Explanation:

First, convert km/hr to m/s:

(75 km/hr)(1000m/1km)(1hr/60min)(1min/60s) = 20.83 m/s

Then, divide 20.93 m/s by 5.0s

(20.93 m/s) / (5.0s) = 4.186 m/s^2

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Which statement is true of the electric field at a distant from the source of charge?
iVinArrow [24]

Answer:

A negative charge, if free to move in an electric field, will move from a low potential point to a high potential point. To move a positive charge against the electric field, work has to be done by you or a force external to the field.

Explanation:

Mark as Brainliest plz!!!

8 0
2 years ago
In a game of angry birds you launch a bird with an angle of 53 degrees to horizontal. Unfortunatly, its not a good shot and the
Alisiya [41]

Answer:

The maximum height covered is 3.25 m.

The horizontal distance covered is 9.81 m.

The total time in the air is 1.63 seconds.

Explanation:

The launch speed, u_0= 10 m/s.

Angle of launch with the horizontal, \theta = 53 ^{\circ}

So, the vertical component of the initial velocity,

u_0\sin\theta=10 \sin 53 ^{\circ}\cdots(i).

The horizontal component of the initial velocity,

u_0\cos\theta=10 \cos 53 ^{\circ}

Let, t be the time of flight, to the horizontal distance covered

D=10 \cos (53 ^{\circ})t\cdots(ii).

Not, applying the equation of motion in the vertical direction.

s= ut +\frac 1 2 at^2

Where s is the displacement in time t, u is the initial velocity and a is the acceleration.

In this case, u =10 \sin 53 ^{\circ} (from equation (i), s=0 (as the final height is same as the launch height) and a = -9.81 m/s^2 (negative sign is due to the downward direction).

\Rightarrow 0 = 10 (\sin 53 ^{\circ})t-\frac 1 2 (9.81)t^2

\Rightarrow t= \frac {2\times 10 (\sin 53 ^{\circ})}{9.81}=1.63 seconds.

So, the total time in the air is 1.63 seconds.

From equation (i),

Total horizontal distance covered is

D=10 \cos (53 ^{\circ})\times 1.63 = 9.81 m.

Now, for the maximum height, H, applying the equation of motion as

v^2=u^2+2as

Here, v is the final velocity and v=0 (at the maximum height), and h=H.

So, 0^2=(10 \sin 53 ^{\circ})^2-2(9.81)H

\Rightarrow H = \frac {(10 \sin 53 ^{\circ})^2}{2\times 9.81}

\Rightarrow H = 3.25 m.

Hence, the maximum height covered is 3.25 m.

8 0
3 years ago
Assuming typical speeds of 8.5 km/s and 5.5 km/s for P and S waves, respectively, how far away did the earthquake occur if a par
taurus [48]

Answer:

The earthquake occurred at a distance of 1122 km

Explanation:

Given;

speed of the P wave, v₁ = 8.5 km/s

speed of the S wave, v₂ =  5.5 km/s

The distance traveled by both waves is the same and it is given as;

Δx = v₁t₁ = v₂t₂

let the time taken by the wave with greater speed = t₁

then, the time taken by the wave with smaller speed, t₂ = t₁ + 1.2 min, since it is slower.

v₁t₁ = v₂t₂

v₁t₁ = v₂(t₁ + 1.2 min)

v₁t₁ = v₂(t₁ + 72 s)

v₁t₁ = v₂t₁ + 72v₂

v₁t₁ - v₂t₁ = 72v₂

t₁(v₁ - v₂) = 72v₂

t_1 = \frac{72v_2}{v_1-v_2}\\\\t_1 =   \frac{72*5.5}{8.5-5.5}\\\\t_1 = 132 \ s

The distance traveled is given by;

Δx = v₁t₁

Δx = (8.5)(132)

Δx = 1122 km

Therefore, the earthquake occurred at a distance of 1122 km

4 0
3 years ago
As denizens of the surface of a spinning planet, we are always in uniform circular motion. Imagine you are in Nairobi (on the Ea
Klio2033 [76]

Answer:

6.02 radian

Explanation:

At 12 noon position is zero radian on Monday

at 12 midnight the position is π/2

at 11 am position on Tuesday is 2π/24 x 23 ( after 23 hours ) = 6..02 radian.

7 0
3 years ago
Can someone please help me ASAP please!
natulia [17]
It would be D (hoping the picture you posted is what you needed help with
8 0
3 years ago
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