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LUCKY_DIMON [66]
3 years ago
12

Which statement is true about vectors?

Physics
1 answer:
Over [174]3 years ago
3 0

Answer:

Vectors are defined as quantities which are fully described by both their magnitude and direction. By definition, a vector has a direction associated with it.

Explanation:

is this what you want?

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A particle moves along a straight line and its position at time t is given by
Alisiya [41]

(A) Use interval notation to indicate the time interval or union of time intervals when the particle is moving forward and backward

the particle is speeding up

<span>if  a(t) >0, it means 12t-48>0, and t >4,  I = ]4, infinity[</span>

the particle is speeding down

<span>if  a(t) < 0, it means 12t-48>0, and t < 4,  I = ] -infinity, 4[</span>

 

when the particle is moving forward and backward

that is depending of the sign of v(t)

moving forward if v(t)> 0

<span>moving backward if v(t)<  0</span>

but v(t) = 6t² - 48 t +90, this quadratic equation has exactly two solutions, t =1, and t =3

<span>so  after taking care the sign of </span>6t² - 48 t +90,

 

<span>v(t)> o when  t is in I = ] –inf, 1[ U ]4, inf[,the particle is moving forward</span>

<span>v(t) < o when  t is in I = ]1, 4[,the particle is moving backward</span>

<span> </span>

3 0
3 years ago
The kicker now kicks the ball with the same speed as in the number of 4,but at 60.0°from the horizontal or 30.0° from the vertic
Shkiper50 [21]

Answer:

i) 0.7

ii) 1.39

iii) 0.6

Next time, when compiling a Physics question, ensure you put the unit of each measurement.

Explanation:

i) T = time of flight =   \frac{2uSin(A)}{g}

where u = speed = 4, A = 60 and  g = acceleration due to gravity = 10 (It is a constant);

Subsituting the values, we have: T = \frac{2(4)Sin(60)}{10} = 0.7

ii) distance travel = Range =  R = \frac{u^{2}Sin(2A) }{g}

where u = speed = 4, A = 60 and  g = acceleration due to gravity = 10 (It is a constant);

Subsituting values, we have: R = \frac{4^{2}Sin(2*60) }{10} = 1.39

iii) Maximum Height = H = \frac{u^{2}(Sin(A))^{2}  }{2g}

where u = speed = 4, A = 60 and  g = acceleration due to gravity = 10 (It is a constant);

Subsituting values, we have: \frac{4^{2}(Sin60)^{2}  }{2*10} = 0.6

4 0
3 years ago
An object, initially at rest, is subject to an acceleration of 34 m/s^2. How long will it take for that object to reach 3400m ?
Norma-Jean [14]
Vf^2 = Vi^2 + 2ad
a= 34 m/s^2
Vi = 0 m/s
d = 3400m

Vf = 480.83 m/s

a=v/t
t=v/a
t=480.83/34
t=14.142 s
6 0
3 years ago
Explain why frog will not look green under the red light?
IrinaVladis [17]
A frog can be many different colours. It appears green under normal 'white' light because it absorbs all the other colours in the light's spectrum apart from green. It reflects the green light back and that is picked up by your eye.

If the light is red, there is no green in the spectrum of the light, only red. So, the red light will be absorbed and there is no green to be reflected back for you to see. Therefore, the frog will not look green.
8 0
3 years ago
Calculate the linear acceleration (in m/s2) of a car, the 0.310 m radius tires of which have an angular acceleration of 15.0 rad
love history [14]

Answer:

a) The linear acceleration of the car is 4.65\,\frac{m}{s^{2}}, b) The tires did 7.46 revolutions in 2.50 seconds from rest.

Explanation:

a) A tire experiments a general plane motion, which is the sum of rotation and translation. The linear acceleration experimented by the car corresponds to the linear acceleration at the center of the tire with respect to the point of contact between tire and ground, whose magnitude is described by the following formula measured in meters per square second:

\| \vec a \| = \sqrt{a_{r}^{2} + a_{t}^{2}}

Where:

a_{r} - Magnitude of the radial acceleration, measured in meters per square second.

a_{t} - Magnitude of the tangent acceleration, measured in meters per square second.

Let suppose that tire is moving on a horizontal ground, since radius of curvature is too big, then radial acceleration tends to be zero. So that:

\| \vec a \| = a_{t}

\| \vec a \| = r \cdot \alpha

Where:

\alpha - Angular acceleration, measured in radians per square second.

r - Radius of rotation (Radius of a tire), measured in meters.

Given that \alpha = 15\,\frac{rad}{s^{2}} and r = 0.31\,m. The linear acceleration experimented by the car is:

\| \vec a \| = (0.31\,m)\cdot \left(15\,\frac{rad}{s^{2}} \right)

\| \vec a \| = 4.65\,\frac{m}{s^{2}}

The linear acceleration of the car is 4.65\,\frac{m}{s^{2}}.

b) Assuming that angular acceleration is constant, the following kinematic equation is used:

\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}

Where:

\theta - Final angular position, measured in radians.

\theta_{o} - Initial angular position, measured in radians.

\omega_{o} - Initial angular speed, measured in radians per second.

\alpha - Angular acceleration, measured in radians per square second.

t - Time, measured in seconds.

If \theta_{o} = 0\,rad, \omega_{o} = 0\,\frac{rad}{s}, \alpha = 15\,\frac{rad}{s^{2}}, the final angular position is:

\theta = 0\,rad + \left(0\,\frac{rad}{s}\right)\cdot (2.50\,s) + \frac{1}{2}\cdot \left(15\,\frac{rad}{s^{2}}\right)\cdot (2.50\,s)^{2}

\theta = 46.875\,rad

Let convert this outcome into revolutions: (1 revolution is equal to 2π radians)

\theta = 7.46\,rev

The tires did 7.46 revolutions in 2.50 seconds from rest.

3 0
4 years ago
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