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3 years ago

8

A car traveling at 26.8m/s hits the brakes to slow down before turning into a neighborhood. Determine how fast it is going after

slowing down at a rate of -8.2m/s2 for 14.3seconds.

1 answer:

AURORKA [14]3 years ago

6 0

Answer:

The speed of the car after the break was applied is 90.46 m/s

Explanation:

Given;

initial velocity of the car, u = 26.8 m/s

accelertation of the car, a = -8.2 m/s²

time of motion, t = 14.3 s

final velocity, v = ?

Apply the following kinematic equation;

v = u + at

v = 26.8 + (-8.2 x 14.3)

v = 26.8 - 117.26

v = -90.46 m/s

magnitude of the final velocity = 90.46 m/s

Therefore, the speed of the car after the break was applied is 90.46 m/s

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Answer:

4 Joules (if you mean 2000 grams)

4000 Joules (if you mean 2000 kilograms)

Explanation:

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The CIA investigated hypnosis as a possible tool for interrogating prisoners. Why did they decide it was unsuitable for that

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D. Hypnosis can make the subjects talk, but they talk only about their childhoods.

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3 years ago

Read 2 more answers

Particle 1 and particle 2 have masses of m1 = 2.2 × 10-8 kg and m2 = 4.8 × 10-8 kg, but they carry the same charge q. The two pa

Lorico [155]

Answer: $r_2=11.81 cm$

Explanation:

Given

$m_1=2.2\times 10^{-8} kg$

$m_2=4.8\times 10^{-8} kg$

same charge on both masses

potential Energy due to Magnetic Field =Kinetic Energy of Particle

$qV=\frac{mv^2}{2}$

$v=\sqrt{\frac{2qV}{m}}$

and we know

Force due to magnetic field will Provide centripetal Force

$qvB=\frac{mv^2}{r}$

$B=\frac{\sqrt{\frac{2Vm}{q}}}{r}$

and B is equal for both particles

thus $\frac{m}{r^2}=constant$

$\frac{m_1}{r_1^2}=\frac{m_2}{r_2^2}$

$r_2^2=\frac{4.8}{2.2}\times 8^2$

$r_2=11.81 cm$

4 0

3 years ago

A 3.0 kg block is pushed by a 14 N force. If µ = 0.6, will the block move?

Anna71 [15]

Answer:

The block will not move.

Explanation:

We'll begin by calculating the frictional force. This can be obtained as follow:

Coefficient of friction (µ) = 0.6

Mass of block (m) = 3 Kg

Acceleration due to gravity (g) = 10 m/s²

Normal reaction (R) = mg = 3 × 10 = 30 N

Frictional force (Fբ) =?

Fբ = µR

Fբ = 0.6 × 30

Fբ = 18 N

From the calculations made above, the frictional force of the block is 18 N. Since the frictional force (i.e 18 N) is bigger than the force applied (i.e 14 N), the block will not move.

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3 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