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3 years ago

It's velocity changes, but its speed remains the same. The true or false

Physics

1 answer:

statuscvo [17]3 years ago

7 0

Answer:

True

Explanation:

Velocity is a vector quantity, which means that it carries both magnitude and direction. Hence when direction of a particle changes, although magnitude (speed) may remain same, it's velocity changes due to direction change. For ex. A particle is m... A particle is moving along x axis with speed 1m/s, it's velocity will be represented as 1i (i represents unit vector along x)

But if it now starts moving along y axis, it's velocity is 1j (j represents unit vector along y axis). Hence velocity changes with direction.

brainllest pls .

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A particle leaves the origin with an initial velocity v → = (3.00iˆ) m/s and a constant acceleration a → = (−1.00iˆ − 0.500jˆ) m

tatiyna

Answer:

the position vector (x,y) will be (1.5 m,-2.25 m) and the velocity vector (vx,vy) will be ( 0 m/s , -1.5 m/s) when x reaches its maximum x coordinate

Explanation:

Since the velocity is related with the acceleration and coordinates through

vx²=v₀x²+2*ax*x

where

vx = velocity in the x direction

v₀x = initial velocity in the x direction = 3 m/s

ax = acceleration in the x direction = −1.00 m/s²

x= coordinates in the x-axis

when x reaches its maximum coordinate , then vx=0

thus

vx²=v₀x²+2*ax*x

0 = (3 m/s)² + 2* (−1.00 m/s²)*x

x= 1.5 m

also for the time t

vx = v₀x + ax*t → t= (vx-v₀x)/ax = (0- 3 m/s)/ (−1.00 m/s²) = 3 seconds

for the y coordinates

y = y₀+v₀y*t + 1/2 ay*t²

where

v₀y = initial velocity in the y direction = 0 m/s

ay = acceleration in the x direction = −0.5 m/s²

y= coordinates in the y-axis

y₀= initial coordinate in the y-axis =0

then since y₀=0 and v₀y=0

y = 1/2*ay*t²

y = 1/2*ay*t² = 1/2*(−0.5 m/s²)*(3 s)² = -2.25 m

and

vy=v₀y+ ay*t= 0+(−0.5 m/s²)*(3 s)= (-1.5 m/s)

therefore the position vector (x,y) will be (1.5 m,-2.25 m)

and the velocity vector (vx,vy) will be ( 0 m/s , -1.5 m/s)

7 0

3 years ago

The Voltage difference of a circuit is 16V and the resistance is 120 ohms. What is the current in the circuit? Show me your work

Ganezh [65]

Answer:

0.133 A

Explanation:

Given:

The voltage difference is, $V=16\ V$

The resistance of the circuit is, $R=120\ ohms$

Now, as per Ohm's Law, the potential difference across two points in a circuit is directly proportional to the current supplied in the circuit. The constant of proportionality is the resistance of the circuit.

Hence, the mathematical statement of Ohm's law is given as:

$V=IR$ Where $I$ is the current in the circuit.

Now, plug in 16 for 'V' and 120 for 'R' and solve for 'I'. This gives,

$16=120I\\I=\frac{16}{120}\\I=0.133\ A$

Therefore, the current in the circuit is 0.133 amperes.

7 0

3 years ago

Explain your thinking. What definition, rule, or reasoning did you use to decide

Vladimir79 [104]

Answer:

Explanation:

A Solid is one of the three main states of matter. By definition a solid is something whose molecules are so densely packed together that it allows it to keep its shape. Therefore if the item has a definite shape that does not change on it's own it is a solid. As opposed to a liquid which will take and fill the shape of it's container.

4 0

3 years ago

Which of the following is a noncontact force?

MrRa [10]

Answer:

Gravity between you and the sun

4 0

3 years ago

2. Am 80.0 kg astronaut is training for accelerations that he will experience upon re-entry to earth’s gravity from space. He is

AlexFokin [52]

Answer:

a) v = 26.2 m / s, b) acceleration is radial, a = 27.4 m / s², c) a = 2.8 g and

d) a = - 8.73 10⁻² m / s², τ = 1.09 10⁴ N m

Explanation:

a) For this exercise we can use the relationships between rotational and linear motion

v = w r

let's reduce the magnitudes to the SI system

w = 10 rpm (2pi rad / 1 rev) 1 min / 60s) = 1,047 rad / s

r = 25.0 m

let's calculate

v = 1.047 25.0

v = 26.2 m / s

b) When the body is rotating at constant speed, the relationship must be perpendicular to the speed, therefore the direction of acceleration is radial, that is, towards the center of the circle and its magnitude is

a = v² / r

a = 26.2²/25

a = 27.4 m / s²

c) Let's look for the relationship between the centripetal acceleration and the acceleration due to gravity

a / g = 27.4 / 9.8

a / g = 2.8

a = 2.8 g

d) let's find the deceleration and torque to stop the centripette in 5 min

t = 5 min (60 s / 1min) = 300 s

let's use the rotational kinematics relations

w = w₀ + α t

initial angular velocity is wo = 1,047 rad / s and the final as is stop do w = 0

α = - w₀ / t

α = - 1,047 / 300

α = -3.49 10⁻³ rad / s²

angular and linear are related

a = α r

a = -3.49 10⁻³ 25

a = - 8.73 10⁻² m / s²

the negative sign indicates that the acceleration is stopping the movement

torque is

τ = F r

The force can be found with Newton's second law

F = m a

we substitute

τ = m a r

τ = 5000.0 8.73 10⁻² 25

τ = 1.09 10⁴ N m

7 0

3 years ago