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IrinaVladis [17]

2 years ago

15

An object is moving along the x axis. At t = 5.6 s, the object is at x = +3.0 m and has a velocity of +5.7 m/s. At t = 8.5 s, it

is at x = +9.0 m and its velocity is -1.0 m/s. Find the average acceleration between t = 5.6 s and t = 8.5 s.

1 answer:

Debora [2.8K]2 years ago

3 0

The average acceleration between t = 5.6 s and t = 8.5 s is 2.31 m/s²

<h3>What is acceleration?</h3>

Acceleration is defined as the rate change of velocity with time.

acceleration a = (Δv) / (Δt)

An object is moving with initial velocity u =5.7 m/s and its final velocity v= -1.0 m/s.

Time taken for the change in speed, t= 8.5 - 5.6 = 2.9 seconds

The acceleration is given by

a = (-1 - 5.7)/ 2.9

a = - 2.31 m/s²

|a | = 2.31 m/s²

Thus, the object's acceleration is 2.31 m/s²

Learn more about acceleration.

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Consider the following four objects: a hoop, a flat disk, a solid sphere, and a hollow sphere. Each of the objects has mass M an

Mumz [18]

Answer:

The hoop

Explanation:

We need to define the moment of inertia of the different objects, that is,

DISK:

$I_{disk} = \frac{1}{2} mR^2$

HOOP:

$I_{hoop} = mR^2$

SOLID SPHERE:

$I_{ss} = \frac{2}{5}mR^2$

HOLLOW SPHERE

$I_{hs} = \frac{2}{3}mR^2$

If we have the same acceleration for a Torque applied, then

$mR^2>\frac{2}{3}mR^2>\frac{1}{2} mR^2>\frac{2}{5}mR^2$

$I_{hoop}>I_{hs} >I_{disk}>I_{ss}$

The greatest momement of inertia is for the hoop, therefore will require the largest torque to give the same acceleration

4 0

3 years ago

A cruise ship made a trip to Guam and back. The trip there took 12 hours and the trip

Alecsey [184]

Answer:

The average speed is 8.0 km/h

Explanation:

We use the data for the return journey to calculate the distance travelled using the constant velocity equation:

s = v t =6*4=24km

Note I didn't change any units so the answer comes out in kilometres.

Now use the distance and time taken to travel to Guam to find the average speed:

v=st=24/3=8.0km/h

6 0

3 years ago

A 6.9-kg wheel with geometric radius m has radius of gyration computed about its mass center given by m. A massless bar at angle

Vilka [71]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The value is $\alpha =2.538 \ rad/s^2$

Explanation:

From the question we are told that

The mass of the wheel is m = 6.9 kg

The radius is $r = 0.69 \ m$

The radius of gyration is $k_G = 0.4\ m$

The angle is $\theta = 47^o$

The force which the massless bar is subjected to $F = 22.5 \ N$

Generally given that the wheels rolls without slipping on the flat stationary ground surface, it implies that point A is the center of rotation.

Generally the moment of inertia about A is mathematically represented as

$I_a = I_G + M* r^2$

Here $I_G$ is the moment of inertia about G with respect to the radius of gyration which is mathematically represented as

$I_G = M * k_G$

=> $I_a = k_G* M + M* r^2$

=> $I_a =0.4 * 6.9 + 6.9 * 0.69^2$

=> $I_a =6.045 \ kg \cdot m^2$

Generally the torque experienced by the wheel is mathematically represented as

$\tau = F * cos (47)$

=> $\tau = 22.5 * cos (47)$

=> $\tau = 15.34 \ kg \cdot m^2 \cdot s^{-2}$

Generally this torque is also mathematically represented as

$\tau = I_a * \alpha$

=> $15.34 = 6.045 * \alpha$

=> $\alpha =2.538 \ rad/s^2$

4 0

3 years ago

An airplane is flying 340 km/hr at 12o east of north. the wind is blowing 40 km/hr at 34o south of east. what is the plane's act

jok3333 [9.3K]

Define an x-y coordinate system such that
The positive x-axis = the eastern direction, with unit vector $\hat{i}$ .
The positive y-axis = the northern direction, with unit vector $\hat{j}$ .

The airplane flies at 340 km/h at 12° east of north. Its velocity vector is
$\vec{v}_{1} = 340(sin(15^{o})\hat{i} + cos(15^{o})\hat{j} ) = 88\hat{i} + 328.4\hat{j}$

The wind blows at 40 km/h in the direction 34° south of east. Its velocity vector is
$\vec{v}_{2} =40(cos(34^{o})\hat{i} - sin(24^{o})]\hat{j}) = 33.1615\hat{i} -22.3677\hat{j})$

The plane's actual velocity is the vector sum of the two velocities. It is
$\vec{v}=\vec{v}_{1}+\vec{v}_{2} = 121.1615\hat{i}+306.0473\hat{j}$

The magnitude of the actual velocity is
v = √(121.1615² + 306.0473²) = 329.158 km/h

The angle that the velocity makes north of east is
tan⁻¹ (306.04733/121.1615) = 21.6°

Answer:
The actual velocity is 329.2 km/h at 21.6° north of east.

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3 years ago

The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with acceleration of −5.2

sergejj [24]

Answer:

The car strikes the tree with a final speed of 4.165 m/s

The acceleration need to be of -5.19 m/seg2 to avoid collision by 0.5m

Explanation:

First we need to calculate the initial speed $V_{0}$

$x=V_{0} *t+\frac{1}{2} *a*(t^{2} )\\62.5m=V_{0} *4.15s+\frac{1}{2} *-5.25\frac{m}{s^{2} } *(4.15^{2} )\\V_{0}=25.953\frac{m}{s}$

Once we have the initial speed, we can isolate the final speed from following equation:

$V_{f} =V_{0} +a*t$ $V_{f}= 4.165 \frac{m}{s}$

Then we can calculate the aceleration where the car stops 0.5 m before striking the tree.

To do that, we replace 62 m in the first formula, as follows:

$x=V_{0} *t+\frac{1}{2} *a*(t^{2} )\\62m=25.953\frac{m}{s}*4.15s+\frac{1}{2} *-a\frac{m}{s^{2} } *(4.15^{2} )\\a=-5.19\frac{m}{s^{2} }$

3 0

3 years ago

Read 2 more answers