Choose the more reactive nonmetal. P or O

Bats are extremely adept at catching insects in midair. If a 81.5-g bat flying in one direction at 7.21 m/s catches a 8.11-g ins

aivan3 [116]

You can look that up on google

7 0

2 years ago

The electrons can be modeled as forming a uniform shell of negative charge. What net electric field do they produce at the locat

Nonamiya [84]

Answer:

E=0

Explanation:

The electric field at the centre of the shell is zero because total enclosed charge in the nucleus is zero

7 0

2 years ago

Which statement best describes how electric fields and magnetic fields are alike?

gladu [14]

<span>Electric field repulsive for objects of like charge and attractive for opposite type of charges and for a magnet you can say that like poles repel and unlike attracts so D makes sense</span>

3 0

2 years ago

A marble column of cross-sectional area 1.6 m^2 supports a mass of 26600 kg. The elastic modulus for marble is 5.0 times 10^10 N

vodka [1.7K]

Answer:

Δ L = 2.57 x 10⁻⁵ m

Explanation:

given,

cross sectional area = 1.6 m²

Mass of column = 26600 Kg

Elastic modulus, E = 5 x 10¹⁰ N/m²

height = 7.9 m

Weight of the column = 26600 x 9.8

= 260680 N

we know,

Young's modulus= $\dfrac{stress}{strain}$

stress = $\dfrac{P}{A}$

= $\dfrac{260680}{1.6}$

= 162925

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

now,

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

$\Delta L = \dfrac{162925}{Y}\times L$

$\Delta L = \dfrac{162925}{5 \times 10^10}\times 7.9$

Δ L = 2.57 x 10⁻⁵ m

The column is shortened by Δ L = 2.57 x 10⁻⁵ m

3 0

2 years ago

A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.76 m/s2 for t1 = 20 s. At that point the

alex41 [277]

Answer:

(a) $v_1 = a_1t_1 = 1.76 t_1$

(b) It won't hit

(c) 110 m

Explanation:

(a) the car velocity is the initial velocity (at rest so 0) plus product of acceleration and time t1

$v_1 = v_0 + a_1t_1 = 0 +1.76t_1 = 1.76t_1$

(b) The velocity of the car before the driver begins braking is

$v_1 = 1.76*20 = 35.2m/s$

The driver brakes hard and come to rest for t2 = 5s. This means the deceleration of the driver during braking process is

$a_2 = \frac{\Delta v_2}{\Delta t_2} = \frac{v_2 - v_1}{t_2} = \frac{0 - 35.2}{5} = -7.04 m/s^2$

We can use the following equation of motion to calculate how far the car has travel since braking to stop

$s_2 = v_1t_2 + a_2t_2^2/2$

$s_2 = 35.2*5 - 7.04*5^2/2 = 88 m$

Also the distance from start to where the driver starts braking is

$s_1 = a_1t_1^2/2 = 1.76*20^2/2 = 352$

So the total distance from rest to stop is 352 + 88 = 440 m < 550 m so the car won't hit the limb

(c) The distance from the limb to where the car stops is 550 - 440 = 110 m

8 0

2 years ago