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2 years ago

Can there be situation when velocity is constant but speed is not

Physics

2 answers:

klasskru [66]2 years ago

4 0

Answer:

no I don’t think there can be so my answer is No.

Okay then yes sorry that I must have gotten it wrong before.

Explanation:

sweet-ann [11.9K]2 years ago

4 0

Answer:

yes

Explanation:

as speed is a scalar quantity and velocity is a vector quantity. Which means speed only depends on the magnitude at which it travels but velocity depends on the speed at which travel and also the direction to where the object travels.

pls mark as brainliest

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An astronaut has left the International Space Station to test a new space scooter. Her partner measures the following velocity c

KiRa [710]

Acceleration can be defined as the change of speed in an instant of time, that is

$a = \frac{v_f-v_i}{\Delta t}$

Here,

$v_f =$ Final velocity

$v_i =$ Initial velocity

$\Delta t =$ Change in time

From this expression we will calculate the requested values replacing the variables in each of the given terms

PART A) The values under this condition are:

$v_f = 5.3m/s$

$v_i = 15m/s$

$\Delta t = 10.4s$

Replacing,

$a = \frac{5.3m/s-(15m/s)}{10.4s}$

$a = -0.93m/s^2$

PART B ) The values under this condition are:

$v_f = -15m/s$

$v_i = -5.3m/s$

$\Delta t = 10.4$

Replacing,

$a = \frac{-(15m/s)-(-5.3m/s)}{10.4s}$

$a = -0.93m/s^2$

Therefore the acceleration in the second time interval is $-0.93m/s^2$

PART C) The values under this condition are:

$v_f = -15m/s$

$v_i = 15m/s$

$\Delta t = 10.4$

Replacing,

$a = \frac{-15m/s-(15m/s)}{10.4s}$

$a = -2.9m/s^2$

6 0

3 years ago

. How far apart must two point charges of 75.0 nC (typical of static electricity) be to have a force of 1.00 N between them?

o-na [289]

Answer:

0.00712 m

Explanation:

Given:

Charge on first particle (q₁) = 75 nC = $75\times 10^{-9}\ C$

Charge on second particle (q₂) = 75 nC = $75\times 10^{-9}\ C$

Force (F) = 1.00 N

Separation (d) = ?

The magnitude of force is given by Coulomb's law which states that, the magnitude of force acting between two charged particles separated by a distance is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them.

Therefore, the magnitude of force is given as:

$F=\dfrac{k|q_1||q_2|}{d^2}$

Where, $k=9\times 10^9\ N\cdot m^2/ C^2$ is the coulomb's constant.

Plug in the given values and solve for 'd'. This gives,

$1.00=\frac{9\times 10^9\times 75.0\times 10^{-9}\times 75.0\times 10^{-9} }{d^2}\\\\d^2=\frac{9\times 10^9\times 75.0\times 10^{-9}\times 75.0\times 10^{-9} }{1.00}\\\\d=\sqrt{\frac{9\times 10^9\times 75.0\times 10^{-9}\times 75.0\times 10^{-9} }{1.00}}\\\\d=0.00712\ m$

Therefore, the distance between the charges is 0.00712 m.

6 0

3 years ago

When hung from an ideal spring with spring constant k = 1.5 N/m, it bounces up and down with some frequency ω, if you stop the b

Umnica [9.8K]

Answer:

L = ¼ k g / m

Explanation:

This is an interesting exercise, in the first case the spring bounces under its own weight and in the second it oscillates under its own weight.

The first case angular velocity, spring mass system is

w₁² = k / m

The second case the angular velocity is

w₂² = L / g

They tell us

w₂ = ½ w₁

Let's replace and calculate

√ (L / g) = ½ √ (k / m)

L / g = ¼ k / m

L = ¼ k g / m

7 0

3 years ago

What force causes a rolling ball to slow down and stop

MAXImum [283]

Newtons first law of motion or friction

3 0

3 years ago

Read 2 more answers

HELP ME ASAP PLEASE

padilas [110]

Answer:

( 1000 × 4 = 4,000) (800×3= 2400) (800×2=1600) the answer is 1600 hope it helps

6 0

3 years ago