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maxonik [38]
3 years ago
9

Please help will give brainliest

Physics
1 answer:
andreyandreev [35.5K]3 years ago
4 0
The answer is A.
It will decay to give you Thorium and Helium.

Please give me Brainliest.
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After a projectile is fired into the air, what is the magnitude of the acceleration
sergiy2304 [10]

Answer: Option A;  9.8 m/s^2

Explanation:

When an object is in the air, and there is no air resistance acting on the object, the only force that will act on the object is the gravitational force (on the vertical axis).

Then, if the only force acting on the object is the gravitational force, the acceleration of the object will be equal to the gravitational acceleration.

We know that the gravitational acceleration is equal to:

g = 9.8m/s^2

Then the acceleration on the vertical axis will be equal to:

a(t) = 9.8m/s^2

The correct option is the first one:

A. 9.8 m/s^2

7 0
3 years ago
¿Qué es la velocidad?
egoroff_w [7]

Velocidad-es una cantidad vectorial física; Tanto la magnitud como la dirección son necesarias para definirlo.

5 0
3 years ago
How much heat is released when 432 g of water cools down from 71'c to 18'c?
maria [59]
The heat released by the water when it cools down by a temperature difference \Delta T is
Q=mC_s \Delta T
where
m=432 g is the mass of the water
C_s = 4.18 J/g^{\circ}C is the specific heat capacity of water
\Delta T =71^{\circ}C-18^{\circ}C=53^{\circ} is the decrease of temperature of the water

Plugging the numbers into the equation, we find
Q=(432 g)(4.18 J/g^{\circ}C)(53^{\circ}C)=9.57 \cdot 10^4 J
and this is the amount of heat released by the water.
7 0
3 years ago
Peg P is driven by the forked link OA along the path described by r = eu, where r is in meters. When u = p4 rad, the link has an
8_murik_8 [283]

Answer:

The transverse component of acceleration is 26.32 m/s^2 where as radial the component of acceleration is 8.77 m/s^2

Explanation:

As per the given data

u=π/4 rad

ω=u'=2 rad/s

α=u''=4 rad/s

r=e^u

So the transverse component of acceleration are given as

a_{\theta}=(ru''+2r'u')\\

Here

r=e^u\\r=e^{\pi/4}\\r=2.1932 m

r'=e^u.u'\\r'=2.1932 \times 2\\r'=4.3864 m

So

a_{\theta}=(ru''+2r'u')\\a_{\theta}=(2.1932\times 4+2\times 4.3864 \times 2)\\a_{\theta}=26.32 m/s\\

The transverse component of acceleration is 26.32 m/s^2

The radial component is given as

a_r=r''-r\theta'^2

Here

r''=e^u.u'^2+e^u u''\\r''=2.1932 \times (2)^2+2.1932\times 4\\r''=17.5456 m

So

a_r=r''-ru'^2\\a_r=17.5456-2.1932\times (2)^2\\a_r=8.7728 m/s^2

The radial component of acceleration is 8.77 m/s^2

6 0
3 years ago
Covalent bonds are formed between BLANK
ololo11 [35]
Covalent bonds are formed through an electrostatic attraction between two oppositely charged ions. Hope this Helps :)
6 0
3 years ago
Read 2 more answers
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