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3 years ago

Plzzzzzz help :(((((((((((

Physics

2 answers:

OlgaM077 [116]3 years ago

7 0

Answer:

4. a) F

b) E

c) B

d) D

Hope it helps

RoseWind [281]3 years ago

6 0

F E B D, I think this is right...

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Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.20×10^6 N, one at an angle 14.0∘ west of north,

laila [671]

Answer:

1.45544 J

Explanation:

See attachment

5 0

3 years ago

Photoelectrons with a maximum speed of 8.00 • 106 m/sec are ejected froma surface in the presence of light with a frequency of 6

Andru [333]

The kinetic energy is given by:

$K=\frac{1}{2}mv^2$

We know the mass and the maximum speed, plugging their values in the expression above we have:

$\begin{gathered} K=\frac{1}{2}(9.1\times10^{-31})(8\times10^6)^2 \\ K=2.91\times10^{-17}\text{ J} \end{gathered}$

Therefore, the answer is d.

5 0

1 year ago

At a distance of 11 cm from a presumably isotropic, radioactive source, a pair of students measure 65 cps (cps = counts per seco

Alborosie

To solve the problem, it is necessary the concepts related to the definition of area in a sphere, and the proportionality of the counts per second between the two distances.

The area with a certain radius and the number of counts per second is proportional to another with a greater or lesser radius, in other words,

$A_1*m=M*A_2$

$A_i =Area$

M,m = Counts per second

Our radios are given by

$r_1 = 11cm$

$R_2 = 20cm$

$m = 65cps$

Therefore replacing we have that,

$A_1*m=M*A_2$

$4\pi r_1^2*m = M * 4\pi R_2^2 M$

$r^2*m=MR^2$

$M = \frac{m*r^2}{R^2}$

$M = \frac{65*11^2}{20^2}$

$M = 19.6625cps$

Therefore the number of counts expect at a distance of 20 cm is 19.66cps

7 0

3 years ago

The diffusion rate for a solute is 4.0 x 10^-11 kg/s in a solvent- filled channel that has a cross-sectional area of 0.50 cm^2 a

zlopas [31]

Answer:

$s = 9.6\times 10^{-12}kg/s$

Explanation:

Given:

Solute Diffusion rate = 4.0 × 10⁻¹¹ kg/s

Area of cross-section = 0.50 cm²

Length of channel =0.25 cm

Now for the new channel

Area of cross-section = 0.30 cm²

Length of channel =0.10 cm

let the Solute Diffusion rate of new channel = s

now equating the diffusion rate per unit volume for both the channels

$\frac{4\times 10^{-11}}{0.50\times 0.25}=\frac{s}{0.30\times 0.10}$

thus,

$s = 9.6\times 10^{-12}kg/s$

7 0

3 years ago

To pull an old stump out of the ground, you and a friend tie two ropes to the stump. You pull on it with a force of 500 N to the

zhannawk [14.2K]

Answer:

C. less than 950 N.

Explanation:

Given that

Force in north direction F₁ = 500 N

Force in the northwest F₂ = 450 N

Lets take resultant force R

The angle between force = θ

θ = 45°

The resultant force R

$R=\sqrt{F_1^2+F_2^2+2F_1F_2cos\theta}$

$R=\sqrt{500^2+450^2+2\times 450\times 500\times cos\theta}$

R= 877.89 N

Therefore resultant force is less than 950 N.

C. less than 950 N

Note- When these two force will act in the same direction then the resultant force will be 950 N.

8 0

3 years ago