Which word means rain? A. cirrus B. nimbus C. stratus D. cumulus

A Block slides down an incline that makes an angle of 30? with the horizontal direction. The coefficient of kinetic friction bet

astraxan [27]

Answer:

Acceleration = 2.35 m/ $s^{2}$

Speed = 8.67 m/s

Explanation:

The coefficient of friction , u =0.3

The angle of incline = 30°

The two forces acting on block are weight and friction.

weight along the incline = mg cos60° = $\frac{mg}{2}$ = 0.5 mg

Friction along incline = umg cos30° = mg $0.3\times \frac{\sqrt{3}}{2}$

Friction along incline = 0.26 mg

Net force acting on the weight = (0.5 - 0.26) mg = 0.24 mg

Acceleration = $\frac{net force}{mass}$ = 0.24 g = 2.35 m/ $s^{2}$

The height of incline = 8 m

Length of the inclined edge = 16 m

$v^{2}=u^{2}+2as$

$v^{2}= 2\times 0.24 \times 9.8\times 16$

v= 8.67 m/s

5 0

3 years ago

An electron moves with a speed of 8.0×106m/s along the -z-axis. It enters a region where there is a uniform magnetic field B = (

Crazy boy [7]

Answer:

Acceleration, $a=9.36\times 10^{18}\ m/s^2$

Explanation:

It is given that,

Speed of electron, $v=8\times 10^6\ m/s$

Charge on an electron, $q=1.6\times 10^{-19}\ C$

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

Magnetic field, $B=5.5i-3.7j$

Magnitude, $|B|=\sqrt{5.5^2+(-3.77)^2}=6.66\ T$

Magnetic force is given by :

$F=qvB$

Also, F = ma

$a=\dfrac{qvB}{m}$

$a=\dfrac{1.6\times 10^{-19}\times 8\times 10^6\times 6.66}{9.1\times 10^{-31}}$

$a=9.36\times 10^{18}\ m/s^2$

So, the acceleration of the electron is $9.36\times 10^{18}\ m/s^2$ . Hence, this is the required solution.

5 0

3 years ago

When a jet lands on an aircraft carrier, a hook on the tail of the plane grabs a wire that quickly brings the plane to a halt be

Veronika [31]

The problem seems to be incomplete because there is no question. However, from the problem description, the logical question is to find he acceleration needed by the jet to land on the airplane carrier. The working equation would be:

2ad = v₂² - v₁²
Since the jet stops, v₂ = 0. Substituting the values:
2(a)(95 m) = 0² - [(240 km/h)(1000 m/1 km)(1h/3600 s)]²
Solving for a,
<em>a = -23.39 m/s² (the negative sign indicates that the jet is decelerating)</em>

8 0

3 years ago

Calculate the kinetic energy of a 750 kg compact car moving 50 m/s.

k0ka [10]

935,500 joules because when we use the KE formula KE=1/2mv^2;
KE=1/2(750)(50)^2
KE=375(2500)
KE=935,500 Joules
Hope it helps

6 0

3 years ago

Assume that a satellite orbits mars 150km above its surface. Given that the mass of mars is 6.485 X 10^23kg, and the radius of m

Kisachek [45]

<span>3598 seconds The orbital period of a satellite is u=GM p = sqrt((4*pi/u)*a^3) Where p = period u = standard gravitational parameter which is GM (gravitational constant multiplied by planet mass). This is a much better figure to use than GM because we know u to a higher level of precision than we know either G or M. After all, we can calculate it from observations of satellites. To illustrate the difference, we know GM for Mars to within 7 significant figures. However, we only know G to within 4 digits. a = semi-major axis of orbit. Since we haven't been given u, but instead have been given the much more inferior value of M, let's calculate u from the gravitational constant and M. So u = 6.674x10^-11 m^3/(kg s^2) * 6.485x10^23 kg = 4.3281x10^13 m^3/s^2 The semi-major axis of the orbit is the altitude of the satellite plus the radius of the planet. So 150000 m + 3.396x10^6 m = 3.546x10^6 m Substitute the known values into the equation for the period. So p = sqrt((4 * pi / u) * a^3) p = sqrt((4 * 3.14159 / 4.3281x10^13 m^3/s^2) * (3.546x10^6 m)^3) p = sqrt((12.56636 / 4.3281x10^13 m^3/s^2) * 4.458782x10^19 m^3) p = sqrt(2.9034357x10^-13 s^2/m^3 * 4.458782x10^19 m^3) p = sqrt(1.2945785x10^7 s^2) p = 3598.025212 s Rounding to 4 significant figures, gives us 3598 seconds.</span>

8 0

3 years ago