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3 years ago

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Physics

1 answer:

baherus [9]3 years ago

8 0

The answer is c to take a pictures with the camera in the metal hall

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If molecules in a substance move FASTER will the TEMPERATURE increase or decrease?

statuscvo [17]

The temperature will increase

4 0

3 years ago

A spacecraft is fueled using hydrazine (N2H4; molecular weight of 32 grams per mole [g/mol]) and carries 1640 kilograms [kg] of

Sauron [17]

Answer:

The value is $t = 689.029 \ hours$

Explanation:

From the question we are told that

The molar mass of hydrazine is $Z = 32 g/mol = \frac{32}{1000} = 0.032 \ kg/mol$

The initial temperature is $T_i = -186 ^o F = (-186-32) *\frac{5}{9} +273.15 = 152\ K$

The final temperature is $T_f = 78 ^o F = (78-32) *\frac{5}{9} +273.15 = 298.7 \ K$

The specific heat capacity is $c_h = 0.099 [kJ/(mol K)] = 0.099 *10^3 J/(mol/K)$

The power available is $P = 300 \ W$

The mass of the fuel is $m = 1640 \ kg$

Generally the number of moles of hydrazine present is

$n = \frac{m}{Z}$

=> $n = \frac{1640}{= 0.032}$

=> $n = 51250 \ mol$

Generally the quantity of heat energy needed is mathematically represented as

$Q = n * c_h * (T_f -T_i)$

=> $Q = 51250 * 0.099 *10^3 * (298.7 - 152)$

=> $Q = 7.441516913 * 10^{8} \ J$

Generally the time taken is mathematically represented as

$t = \frac{Q}{P}$

=> $t = \frac{7.441516913 * 10^{8} }{300}$

=> t = 2480505.6377 s

Converting to hours

$t = \frac{2480505.6377}{3600}$

=> $t = 689.029 \ hours$

6 0

2 years ago

A small object begins a free-fall from a height of =81.5 m at 0=0 s . After τ=2.20 s , a second small object is launched vertica

-BARSIC- [3]

Answer:

33.2 m

Explanation:

For the first object:

y₀ = 81.5 m

v₀ = 0 m/s

a = -9.8 m/s²

t₀ = 0 s

y = y₀ + v₀ t + ½ at²

y = 81.5 − 4.9t²

For the second object:

y₀ = 0 m

v₀ = 40.0 m/s

a = -9.8 m/s²

t₀ = 2.20 s

y = y₀ + v₀ t + ½ at²

y = 40(t−2.2) − 4.9(t−2.2)²

When they meet:

81.5 − 4.9t² = 40(t−2.2) − 4.9(t−2.2)²

81.5 − 4.9t² = 40t − 88 − 4.9 (t² − 4.4t + 4.84)

81.5 − 4.9t² = 40t − 88 − 4.9t² + 21.56t − 23.716

81.5 = 61.56t − 111.716

193.216 = 61.56t

t = 3.139

The position at that time is:

y = 81.5 − 4.9(3.139)²

y = 33.2

7 0

2 years ago

Two cars, both of mass m, collide and stick together. Prior to the collision, one car had been traveling north at a speed 2v, wh

mel-nik [20]

Answer: $u=\frac{v}{2}\sqrt{5-4sin\phi }$

Explanation:

Given

Both cars mass is m

and solving problem in Vertical and horizontal direction

considering + y and +x to be positive and u be the final velocity of system

Conserving Momentum in Vertical direction

$m(2v)+m(-vsin\phi )=2m(ucos\theta )$

$2ucos\theta =v(2-sin\theta )$ ------1

Conserving momentum in x direction

$mvcos\phi =2musin\theta$ -----2

squaring and adding 1 &2

$(2u)^2=(2v-vsin\phi )^2+(vcos\phi )^2$

$4u^2=4v^2+v^2-4v^2sin\phi$

$4u^2=5v^2-4v^2sin\phi$

$u=\frac{v}{2}\sqrt{5-4sin\phi }$

7 0

2 years ago

I need help with this

lord [1]

Answer:

0.832

Explanation:

8.320 x 10 to the negative 1st power is 0.832

7 0

3 years ago