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3 years ago

14

An object accelerates to a velocity of 230 m/s over a time of 2.5 s. The acceleration it experienced was 42 m/s2. What was its i

nitial velocity?

1 answer:

zmey [24]3 years ago

5 0

Answer:

230 = x + 105

x= 125

Explanation:

v = v0 + at

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A 20 kg wagon is pulled along the level ground by a rope inclined at 30 degree above the horizontal. A friction force of 30 N op

Elan Coil [88]

(a) 34.6 N

To solve the problem, we have to analyze the forces acting along the horizontal direction.

We have:

- Forward: the component of the pull parallel to the ground, which is

$F cos \theta$

where

F is the magnitude of the pull

$\theta=30^{\circ}$ is the angle

- Backward: the force of friction, which is

$F_f = 30 N$

So, the equation of motion is

$F cos \theta - F_f = ma$

where

m = 20 kg is the mass of the wagon

a is the acceleration

In this part, the wagon is moving at constant speed, so a =0 and the equation becomes

$F cos \theta - F_f = 0$

Therefore, we can find the pulling force:

$F=\frac{F_f}{cos \theta}=\frac{30}{cos 30}=34.6 N$

(b) 43.9 N

In this case, the acceleration is

$a=0.40 m/s^2$

So, the equation of motion in this case is

$F cos \theta - F_f = ma$

So this time we have to take into account the term (ma).

Using the same data as before:

m = 20 kg

$\theta=30^{\circ}$

$F_f = 30 N$

We find the new magnitude of F:

$F=\frac{ma+F_f}{cos \theta}=\frac{(20)(0.40)+30}{cos 30}=43.9 N$

6 0

3 years ago

A car transports its passengers between 3 buildings. It moves from the first building to the second building, 4.76km away, in a

BigorU [14]

Answer:

1) a = 6.14 km

2) b = 4.69 km

Explanation:

Let the first building be A, second building be B and third building be C.

Now, bearing of A = 4.76 km in a direction 37° north of east

Bearing of B = 69° west of north

Bearing of C = 28° east of south

Thus if this 3 points form a triangle, we will have the following angles;

Angle at point A = 28 + (90 - 37) = 81°

Angle at point B = 28 + (90 - 69) = 49°

Angle at point C = 180 - (81 + 49) = 50°

Now, the distance between second and third building is "a" which is represented by BC in the triangle attached while the given distance of 4.76 represents side AB. Thus;

Using sine rule, we can find "a".

a/sin 81 = 4.76/sin 50

a = 6.14 km

B) distance between first and third building is AB in the triangle depicted by "b".

Similar to the first problem, we will use sine rule again.

b/sin49 = 4.76/sin 50

b = 4.69 km

8 0

3 years ago

A car of mass M = 1000 kg traveling at 50.0 km/hour enters a banked turn covered with ice. The road is banked at an angle θ , an

kupik [55]

The radius of the curved road at the given condition is 54.1 m.

The given parameters:

<em>mass of the car, m = 1000 kg</em>
<em>speed of the car, v = 50 km/h = 13.89 m/s</em>
<em>banking angle, θ = 20⁰</em>

The normal force on the car due to banking curve is calculated as follows;

$Fcos(\theta) = mg$

The horizontal force on the car due to the banking curve is calculated as follows;

$Fsin(\theta) = \frac{mv^2}{r}$

<em>Divide </em><em>the second equation by the first;</em>

$\frac{Fsin(\theta)}{Fcos(\theta) } = \frac{mv^2}{rmg} \\\\tan(\theta) = \frac{v^2}{rg} \\\\r = \frac{v^2}{g \times tan(\theta)} \\\\r = \frac{13.89^2}{9.8 \times tan(20)} \\\\r = 54.1 \ m$

Thus, the radius of the curved road at the given condition is 54.1 m.

Learn more about banking angle here: brainly.com/question/8169892

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2 years ago

You charge an initially uncharged 65.7-mf capacitor through a 39.1-Ï resistor by means of a 9.00-v battery having negligible int

uysha [10]

In a RC-circuit, with the capacitor initially uncharged, when we connect the battery to the circuit the charge on the capacitor starts to increase following the law:
$Q(t) = Q_0 (1-e^{-t/\tau})$
where t is the time, $Q_0 = CV$ is the maximum charge on the capacitor at voltage V, and $\tau = RC$ is the time constant of the circuit.
Using this law, we can answer all the three questions of the problem.

1) Using $R=39.1 \Omega$ and $C= 65.7 mF=65.7\cdot 10^{-3}F$ , the time constant of the circuit is:
$\tau = RC=(39.1 \Omega)(65.7 \cdot 10^{-3}F)=2.57 s$

2) To find the charge on the capacitor at time $t=1.95 \tau$ , we must find before the maximum charge on the capacitor, which is
$Q_0 = CV=(65.7 \cdot 10^{-3}F)(9 V)=0.59 C$
And then, the charge at time $t=1.95 \tau$ is equal to
$Q(1.95 \tau) = Q_0 (1-e^{-t/\tau})=(0.59 C)(1-e^{-1.95})=0.51 C$

3) After a long time (let's say much larger than the time constant of the circuit), the capacitor will be fully charged, this means its charge will be $Q_0 = 0.59 C$ . We can see this also from the previous formule, by using $t=\infty$ :
$Q(t) = Q_0 (1-e^{-\infty})=Q_0(1-0) = 0.59 C$

4 0

3 years ago

After your school's team wins the regional championship, students go to the dorm roof and start setting off fireworks rockets. T

PIT_PIT [208]

Answer:

160 m

Explanation:

The intensity, I, of the sound is inversely proportional to the square of the distance, r, from the source.

$I\propto \dfrac{1}{r^2}$

Hence,

$I_1r_1^2 = I_2r_2^2$

$r_2 = r_1\sqrt{\dfrac{I_1}{I_2}}$

From the question, $I_2$ is half of $I_1$

$r_2 = r_1\sqrt{\dfrac{I_1}{0.5I_1}}$

$r_2 = r_1\sqrt{2}$

$r_2 = 113\text{ m}\sqrt{2} = 160 \text{ m}$

3 0

3 years ago

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