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pochemuha
2 years ago
8

I NEEEEEEEEEEEEEEEEEED Help:(​

Physics
1 answer:
soldier1979 [14.2K]2 years ago
8 0

Answer:

C. Light rays are bent by the water in the glass

Explanation:

That's the way science works and I learned about this before in class.

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Where did beowulf and his warriors take the skull
dedylja [7]
I'm pretty sure Beowulf took the skull to Herot.
6 0
3 years ago
A humanoid skeleton is found buried in the ashes of a volcano that erupted between 10,000 and 12,000 years ago. when scientists
Inga [223]
From reliable sources in the internet, the half-live of carbon-14 is given to be 5,730 years. In a span of 10,000 to 12,000 years, there are almost or little more than 2 half-lives. Thus, there should be
                                           A(t) = A(0)(1/2)^t
where t is the number of half-lives, in this case 2. Thus, only about 1/4 of the original amount will be left. 
5 0
3 years ago
Read 2 more answers
Keeping the mass at 1.0 kg and the velocity at 10.0 m/s, record the magnitude of centripetal acceleration for each given radius
Paha777 [63]

Answer:

The centripetal acceleration for the first radius; 2.0 m = 50 m/s²

The centripetal acceleration for the second radius; 4.0 m = 25 m/s²

The centripetal acceleration for the third radius; 6.0 m = 16.67 m/s²

The centripetal acceleration for the fourth radius; 8.0 m = 12.5 m/s²

The centripetal acceleration for the fifth radius; 10.0 m = 10 m/s²

Explanation:

Given;

mass of the object, m = 1 kg

velocity of the object, v = 10 m/s

different values of the radius, 2.0 m 4.0 m 6.0 m 8.0 m 10.0 m

The centripetal acceleration for the first radius; 2.0 m

a_c = \frac{v^2}{r} \\\\a_c_1= \frac{(10)^2}{2} \\\\a_c_1= 50 \ m/s^2

The centripetal acceleration for the second radius; 4.0 m

a_c_2= \frac{(10)^2}{4} \\\\a_c_2= 25 \ m/s^2

The centripetal acceleration for the third radius; 6.0 m

a_c_3= \frac{(10)^2}{6} \\\\a_c_3= 16.67 \ m/s^2

The centripetal acceleration for the fourth radius; 8.0 m

a_c_4= \frac{(10)^2}{8} \\\\a_c_4= 12.5 \ m/s^2

The centripetal acceleration for the fifth radius; 10.0 m

a_c_5= \frac{(10)^2}{10} \\\\a_c_5= 10 \ m/s^2

6 0
3 years ago
On February 15, 2013, Asteroid 2012 DA14 passed within 17,200 miles [mi] of the surface of the Earth at a relative speed of 7.8
xz_007 [3.2K]

Answer:

The total amount of energy that would have been released if the asteroid hit earth = The kinetic energy of the asteroid = 1.29 × 10¹⁵ J = 1.29 PetaJoules = 1.29 PJ

1 PJ = 10¹⁵ J

Explanation:

Kinetic energy = mv²/2

velocity of the asteroid is given as 7.8 km/s = 7800 m/s

To obtain the mass, we get it from the specific gravity and diameter information given.

Density = specific gravity × 1000 = 3 × 1000 = 3000 kg/m³

But density = mass/volume

So, mass = density × volume.

Taking the informed assumption that the asteroid is a sphere,

Volume = 4πr³/3

Diameter = 30 m, r = D/2 = 15 m

Volume = 4π(15)³/3 = 14137.2 m³

Mass of the asteroid = density × volume = 3000 × 14137.2 = 42411501 kg = 4.24 × 10⁷ kg

Kinetic energy of the asteroid = mv²/2 = (4.24 × 10⁷)(7800²)/2 = 1.29 × 10¹⁵ J

6 0
3 years ago
An electron moving parallel to a uniform electric field increases its speed from 2.0 × 107 m/s to 4.0 × 107 m/s over a distance
jeka94

Answer:

1.8\times 105 N/C

Explanation:

We are given that

u=2\times 10^7 m/s

v=4\times 10^7 m/s

d=1.9 cm=\frac{1.9}{100}=0.019 m

Using 1m=100 cm

We have to find the electric field strength.

v^2-u^2=2as

Using the formula

(4\times 10^7)^2-(2\times 10^7)^2=2a(0.019)

16\times 10^{14}-4\times 10^{14}=0.038a

0.038a=12\times 10^{14}

a=\frac{12}{0.038}\times 10^{14}=3.16\times 10^{16}m/s^2

q=1.6\times 10^{-19} C

Mass of electron,m=9.1\times 10^{-31} kg

E=\frac{ma}{q}

Substitute the values

E=\frac{9.1\times 10^{-31}\times 3.16\times 10^{16}}{1.6\times 10^{-19}}

E=1.8\times 105 N/C

7 0
4 years ago
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