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3 years ago

An object that is slowing down in a positive direction must have

Physics

1 answer:

zepelin [54]3 years ago

5 0

Answer:

Positive velocity and negative acceleration

Explanation:

An object moving in the positive direction has a positive velocity.

An object that's slowing down while moving in the positive direction has a negative acceleration.

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A large ant is standing on the middle of a circus tightrope that is stretched with tension T_s. The rope has mass per unit lengt

k0ka [10]

Answer:

$A = \frac{g \lambda^2 \mu}{4\pi^2 T_s}$

Explanation:

As we know that the speed of the wave in string is given as

$v = \sqrt{\frac{T_s}{\mu}}$

now we will have

frequency of the wave is given as

$f = \frac{v}{\lambda}$

$f = \frac{1}{\lambda}\sqrt{\frac{T_s}{\mu}}$

now if the ant will feel weightlessness then we will have

$mg = m\omega^2 A$

so we will have

$A = \frac{g}{\omega^2}$

$A = \frac{g}{4\pi^2 f^2}$

$A = \frac{g \lambda^2 \mu}{4\pi^2 T_s}$

3 0

4 years ago

s A horizontal insulating rod of length 11.0-cm and charge 19 nC is in a plane with a long straight vertical uniform line charge

guajiro [1.7K]

Answer:

F = 2.26 × 10⁻³ N

Explanation:

given,

length of rod = 11 cm

charge = 19 nC

linear charge density = 3.9 x 10⁻⁷ C/m

electric force at 2 cm away.

$E(r) = \dfrac{2K\lambda}{r}$

F = E q

$F= \dfrac{2K\lambda\ q}{L}\int \dfrac{dr}{r}$

integrating from 0.02 to 0.02 + L

$F= \dfrac{2K\lambda\ q}{L}[ln(0.02+L)-ln(0.002)]$

$F= \dfrac{2\times 9 \times 10^9\times 3.9\times 10^{-7}\times 19 \times 10^{-9}}{0.11}[ln(0.02+0.11)-ln(0.002)]$

F = 2.26 × 10⁻³ N

5 0

4 years ago

A parallel-plate capacitor has plates of area A. The plates are initially separated by a distance d, but this distance can be va

ohaa [14]

Answer:

d' = d /2

Explanation:

Given that

Distance = d

Voltage =V

We know that energy in capacitor given as

$U=\dfrac{1}{2}CV^2$

$C=\dfrac{\varepsilon _oA}{d}$

$U=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d}\times V^2$

If energy become double U' = 2 U then d'

$U'=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d'}\times V^2$

$2U=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d'}\times V^2$

$2\times \dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d}\times V^2=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d'}\times V^2$

2 d ' = d

d' = d /2

So the distance between plates will be half on initial distance.

7 0

3 years ago

A dog runs with an initial speed of 1.5 m/s on a floor. It slides to a stop with an acceleration of -0.35 m/s^2. How long does i

Neko [114]

Answer:

3.214 m

Explanation:

Here object is moving in a constant acceleration. Then we can use motion equations to find the total distance

V² = U² + 2as

0 = 1.5² + 2×-0.35 × s

s = 3.214 m

symbols has usual meanings.

6 0

4 years ago

Read 2 more answers

1kg slab of concrete loses 12,000j of heat when it cools from 30 Celsius to 26 Celsius. Determine the specific heat capacity of

OleMash [197]

The specific heat capacity of concrete is $3.0 J/(g^{\circ}C)$

Explanation:

When a certain amount of energy Q is supplied/given off to/from a sample of substance with mass m, the temperature of the substance increases/decreases by an amount $\Delta T$ , according to the equation

$Q=mC_s \Delta T$

where

m is the mass of the substance

$C_s$ is the specific heat capacity of the substance

$\Delta T$ is the change in temperature of the substance

In this problem, we have:

m = 1 kg = 1000 g is the mass of the concrete slab

$Q = -12,000 J$ is the amount of energy lost by the slab

$\Delta T = 30-26= -4^{\circ}C$ is the change in temperature of the slab

Solving the equation for $C_s$ , we find the specific heat capacity of concrete:

$C_s = \frac{Q}{m \Delta T}=\frac{-12,000}{(1000)(-4)}=3.0 J/(g^{\circ}C)$

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

3 0

3 years ago