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Arte-miy333 [17]

3 years ago

15

A cart pulled by a force of 250N at an angle of 35 degrees above the horizontal. The cart accelerates at 1.4 m/s^2. The mass of

the cart, to the nearest whole number, is?

1 answer:

-BARSIC- [3]3 years ago

6 0

<span> 146 kg is the answer to your question hope this helps bro

</span>

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How much work is done on a satellite in a circular orbit about earth?

alexgriva [62]

Answer:

0 J

Explanation:

$W$ = Work done on the satellite in circular orbit about earth by earth

$F$ = Force on gravity on satellite by earth

$d$ = displacement of the satellite

$\theta$ = Angle between the force on gravity and displacement = 90

We know that, Work done is given as

$W = F d Cos\theta\\W = F d Cos90\\W = F d (0)\\W = 0 J$

3 0

3 years ago

Reaction rates are affected by reactant concentrations and temperature. this is accounted for by the ________.

Black_prince [1.1K]

<span>
Reaction rates are affected by reactant concentrations and temperature. this is accounted for by the c</span>ollision model.

-Hope this helps.

7 0

3 years ago

A positively charged non-metal ball A is placed near metal ball B. Prove that if the charge on B is positive but of small magnit

nordsb [41]

Answer:

D

Explanation:

Ball A is a non positively charged non metal while ball B is metal ball.

Given: The ball B positive charge of small magnitude

To prove: Balls will attract each other

IN this condition because of induction negative charges on the sphere B move to the side closer to sphere A,(induction charging) while the positive charges move to the side further away from sphere A. The polarization of charge on B will cause a greater attractive force than the repulsion of the like charges.

Hence the correct answer will be D .

8 0

3 years ago

when light falls obliquely, there is a change in direction as it moves from one medium to another but when it falls perpendicula

exis [7]

Answer:

wait in comments....................

3 0

2 years ago

1. Bone has a Young’s modulus of about

Blababa [14]

#1

As we know that

$Y = \frac{stress}{strain}$

now plug in all data into this

$1.8\times 10^{10} = \frac{1.68 \times 10^8}{strain}$

$strain = 9.33 \times 10^{-3}$

now from the formula of strain

$strain = \frac{\Delta L}{L}$

$9.33 \times 10^{-3} = \frac{\Delta L}{0.54}$

$\Delta L = 5.04 \times 10^{-3} m$

$\Delta L = 5.04 mm$

#2

As we know that

pressure * area = Force

here we know that

$Area = 3.53 \times 11.6 = 40.95 m^2$

$P = 0.2 atm = 0.2 \times 1.01 \times 10^5 = 2.02\times 10^4 Pa$

now force is given as

$F = 40.95 \times (2.02\times 10^4) = 8.27 \times 10^5 N$

#3

As we know that density of water will vary with the height as given below

$\rho = \frac{\rho_0}{1 - \frac{\Delta P}{B}}$

here we know that

$\Delta P = 2600 atm = 2600 \times 1.01 \times 10^5 = 2.63\times 10^8 Pa$

$B = 2.3 \times 10^9 N/m^2$

now density is given as

$\rho = \frac{1050}{1 - \frac{2.63\times 10^8}{2.3 \times 10^9}}$

$\rho = 1185.3 kg/m^3$

#4

as we know that pressure changes with depth as per following equation

$P = P_o + \rho g h$

here we know that

$P = 3 P_0$

now we will have

$3P_0 = P_0 + \rho g h$

$2P_0 = \rho g h$

$2(1.01 \times 10^5) = 1025 (9.81)(h)$

here we will have

$h = 20.1 m$

so it is 20.1 m below the surface

#5

Here net buoyancy force due to water and oil will balance the weight of the block

so here we will have

$mg = \rho_1V_1g + \rho_2V_2g$

$A(0.0476)979 = 922(A)(0.0476 - x) + 1000(A)(x)$

$46.6 = 43.89 - 922x + 1000x$

$x = 3.48 cm$

so it is 3.48 cm below the interface

5 0

3 years ago