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4 years ago

A vector has a length of 8 and rises

Physics

1 answer:

koban [17]4 years ago

3 0

Answer:

0=ten-1(2/8)=14.03

i think i hope this helps

Explanation:

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Find the energy released when there is a decrease of 0.3kg of material in a nuclear reaction

MrRissso [65]

Answer:

E = 2.7 x 10¹⁶ J

Explanation:

The release of energy associated with the mass can be calculated by Einstein's mass-energy relation, as follows:

$E = mc^2$

where,

E = Energy Released = ?

m = mass of material reduced = 0.3 kg

c = speed of light = 3 x 10⁸ m/s

Therefore,

$E = (0.3\ kg)(3\ x\ 10^8\ m/s)^2$

E = 2.7 x 10¹⁶ J

4 0

3 years ago

Answer the following. (a) What is the surface temperature of Betelgeuse, a red giant star in the constellation of Orion, which r

bagirrra123 [75]

Answer:

(a) T = 2987.6 k

(b) T = 19986.2 k

Explanation:

The temperature of a star in terms of peak wavelength can be given by Wein's Displacement Law, which is as follows:

$T = \frac{0.2898\ x\ 10^{-2}\ m.k}{\lambda_{max}}$

where,

T = Radiated surface temperature

$\lambda_{max}$ = peak wavelength

(a)

here,

$\lambda_{max}$ = 970 nm = 9.7 x 10⁻⁷ m

Therefore,

$T = \frac{0.2898\ x\ 10^{-2}\ m.k}{9.7\ x\ 10^{-7}\ m}$

T = 2987.6 k

(b)

here,

$\lambda_{max}$ = 145 nm = 1.45 x 10⁻⁷ m

Therefore,

$T = \frac{0.2898\ x\ 10^{-2}\ m.k}{1.45\ x\ 10^{-7}\ m}$

T = 19986.2 k

6 0

3 years ago

A research group at Dartmouth College has developed a Head Impact Telemetry (HIT) System that can be used to collect data about

Olin [163]

Answer:

6.05 cm

Explanation:

The given equation is

2 aₓ(x-x₀)=( Vₓ²-V₀ₓ²)

The initial head velocity V₀ₓ =11 m/s

The final head velocity Vₓ is 0

The accelerationis given by =1000 m/s²

the stopping distance = x-x₀=?

So we can wind the stopping distance by following formula

2 (-1000)(x-x₀)=[ $0^{2} -11^{2}$ ]

x-x₀=6.05* $10^{-2}$ m

=6.05 cm

3 0

3 years ago

A 975-kg sports car (including driver) crosses the rounded top of a hill at determine (a) the normal force exerted by the road o

iVinArrow [24]

There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill: $v=15 m/s$
- radius of the hill: $r=100 m$

Solution:

(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car $W=mg$ (downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, $m \frac{v^2}{r}$ , so we can write:
$mg-N=m \frac{v^2}{r}$ (1)
By rearranging the equation and substituting the numbers, we find N:
$N=mg-m \frac{v^2}{r}=(975 kg)(9.81 m/s^2)-(975 kg) \frac{(15 m/s)^2}{100 m}=7371 N$

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
$N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N$

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
$mg=m \frac{v^2}{r}$
from which we find
$v= \sqrt{gr}= \sqrt{(9.81 m/s^2)(100 m)}=31.3 m/s$

7 0

4 years ago

Bro what does this mean ? I don’t get this

zhannawk [14.2K]

CO2 is carbon dioxide a colorless gas

3 0

3 years ago

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