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Dmitry_Shevchenko [17]

3 years ago

9

. Calculate the unbalanced force acting on a 0.50 kg toy rocket that accelerates at a constant rate with a velocity of 20 m/s (u

p) to 45 m/s (up) in 0.7 s.

1 answer:

Kruka [31]3 years ago

8 0

Answer:

17.85N

Explanation:

Given data

m= 0.5kg

u= 20m/s

v= 45m/s

t= 0.7second

From

F= mΔv/t

substitute

F=0.5*(45-20)/0.7

F=0.5*25/0.7

F=12.5/0.7

F=17.85

Hence the force is 17.85N

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Answer:

$\text{Vector product}=172.66\ m^2$

Explanation:

Given that,

The magnitude of vector A, $|A|=13\ m$

The magnitude of vector B, $|B|=16\ m$

Scalar product of A and B, $A{\cdot} B=116\ m^2$

The formula for the scalar product is given by :

$A{\cdot} B=|A||B|\cos\theta$

Where, $\theta$ is the angle between A and B.

$\cos\theta=\dfrac{116}{13\times 16}\\\\\theta=\cos^{-1}\left(0.5576\right)\\\\\theta=56.11^{\circ}$

The formula for the vector product is given by :

$A\times B=|A||B|\sin\theta\\\\=13\times 16\times \sin56.11\\\\=172.66\ m^2$

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A college student likes to take her books to class by placing them in a box and pulling the box around behind her. She pulls on

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Answer:

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- Katty queen

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The magnitude of a uniform electric field between two plates is about 1.7 × 106N/C. If the distance

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Answer:

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4 years ago

Read 2 more answers

One end of a metal rod is in contact with a thermal reservoir at 745. K, and the other end is in contact with a thermal reservoi

Masteriza [31]

Answer:

a) $S_1=-9.65}\ J/K$

b) $S_2=71.18\ J/K$

c) 0 J/K

d)S= 61.53 J/K

Explanation:

Given that

T₁ = 745 K

T₂ = 101 K

Q= 7190 J

a)

The entropy change of reservoir 745 K

$S_1=-\dfrac{7190}{745}\ J/K$

Negative sign because heat is leaving.

$S_1=-9.65}\ J/K$

b)

The entropy change of reservoir 101 K

$S_2=\dfrac{7190}{101}\ J/K$

$S_2=71.18\ J/K$

c)

The entropy change of the rod will be zero.

d)

The entropy change of the system

S= S₁ + S₂

S = 71.18 - 9.65 J/K

S= 61.53 J/K

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4 years ago