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3 years ago

Solve 1-2 a,b,c please

Physics

1 answer:

antoniya [11.8K]3 years ago

3 0

B the answer is b in this equation

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Sam, whose mass is 75 kg , takes off across level snow on his jet-powered skis. The skis have a thrust of 160 N and a coefficien

Volgvan

Answer:

top speed = 17.25

Total height = 281.19 m

Explanation:

given data

mass = 75 kg

thrust = 160 N

coefficient of kinetic friction = 0.1

solution

we get here frictional force acting that is

frictional force = $\mu *m*g$ .............1

frictional force = 0.1 × 75 × 9.8

frictional force = 73.5 N

and

Net force acting will be F = 160 - 73.5 N

F = 86.5 N

Acceleration in the First 15 second will be

F = ma .........2

86.5 = 75 × a

a = 1.15 m/s²

and

now After 15 second the velocity will be as

v = u + at ..........3

here u is 0

so v will be

V = 1.15 × 15

v = 17.25

and

now we get travels distance S in 15 s

s = u × t + 0.5 × a × t²

here u is 0

so distance s will be

s = 0.5 × a × t²

s = 0.5 × 1.15 × 15²

s = 129.37 m

and

now acceleration acting is

F = $\mu *m*g$

m a = $\mu *m*g$

a = $\mu* g$

a = - 0.98

here it is negative it mean downward nature of acceleration

and

now we get distance s by this formula

V² - u² = 2 a s

here v velocity is 0 and

u initial velocity is 17.25 m/s

put here value

0 - 17.25² = 2 × (-0.98) × s

solve it we get

s = 151.82 m

Total height is

Total height = 129.37 m + 151.82 m

Total height = 281.19 m

7 0

3 years ago

To practice Problem-Solving Strategy 7.2 Problems Using Mechanical Energy II. The Great Sandini is a 60.0-kg circus performer wh

sp2606 [1]

Answer:

$v = 15.45 m/s$

Explanation:

As per mechanical energy conservation we can say that here since friction is present in the barrel so we will have

Work done by friction force = Loss in mechanical energy

so we will have

$W_f = (U_i + K_i) - (U_f + K_f)$

here we know that

$W_f = F_f . d$

$W_f = 40 \times 4$

$W_f = 160 J$

Initial compression in the spring is given as

$F = kx$

$4400 = 1100 x$

$x = 4 m$

now from above equation

$W_f = (\frac{1}{2}kx^2 + 0) - (mgh + \frac{1}{2}mv^2)$

$160 = (\frac{1}{2}1100(4^2) + 0) - (60 \times 9.8\times 2.50 + \frac{1}{2}(60)v^2)$

$160 = 8800 - 1470 - 30 v^2$

$v = 15.45 m/s$

3 0

3 years ago

A 6.99-g bullet is moving horizontally with a velocity of +341 m/s, where the sign + indicates that it is moving to the right (s

Ratling [72]

Answer:

a). 1.218 m/s

b). R=2.8 $^{-3}$

Explanation:

$m_{bullet}=6.99g*\frac{1kg}{1000g}=6.99x10^{-3}kg$

$v_{bullet}=341\frac{m}{s}$

Momentum of the motion the first part of the motion have a momentum that is:

$P_{1}=m_{bullet}*v_{bullet}$

$P_{1}=6.99x10^{-3}kg*341\frac{m}{s} \\P_{1}=2.3529$

The final momentum is the motion before the action so:

a).

$P_{2}=m_{b1}*v_{fbullet}+(m_{b2}+m_{bullet})*v_{f}}$

$P_{2}=1.202 kg*0.554\frac{m}{s}+(1.523kg+6.99x10^{-3}kg)*v_{f}$

$P_{1}=P_{2}$

$2.529=0.665+(1.5299)*v_{f}\\v_{f}=\frac{1.864}{1.5299}\\v_{f}=1.218 \frac{m}{s}$

b).

kinetic energy

$K=\frac{1}{2}*m*(v)^{2}$

Kinetic energy after

$Ka=\frac{1}{2}*1.202*(0.554)^{2}+\frac{1}{2}*1.523*(1.218)^{2}\\Ka=1.142 J$

Kinetic energy before

$Kb=\frac{1}{2}*mb*(vf)^{2}\\Kb=\frac{1}{2}*6.99x10^{-3}kg*(341)^{2}\\Kb=406.4J$

Ratio = $\frac{Ka}{Kb}$

$R=\frac{1.14}{406.4}\\R=2.8x10^{-3}$

3 0

3 years ago

The speed of the center of the earth as it orbits the sun is 107257 kmph and the absolute angular velocity of the earth about it

Sindrei [870]

Answer with Explanation:

We are given that

Speed , $v_0$ =107257 kmph

Angular velocity, $\omega=7.292\times 10^{-5} rad/s$

Radius of earth,r= $6371 km=6371000 m$

1 km=1000 m

Linear velocity, $v=r\omega=6371000\times 7.292\times 10^{-5}=464.57 m/s$

Linear velocity, $v=464.57\times \frac{18}{5}=1672.46 km/h$

Velocity at point A, $v_A=-vi+v_0 j=-1672.46 i+107257j kmph$

Velocity at point B, $v_B=v_0j-vj=107257j-1672.46j=105584.54j kmph$

Velocity at point C, $v_C=v_0j+vi=1672.46 i+107257j kmph$

Velocity at point D, $v_D=v_0j+vj=107257j+1672.46j=108929.46jkmph$

3 0

3 years ago

We see a full moon by reflected sunlight. How much earlier did the light that enters our eye leave the sun? the earth-moon and e

Tju [1.3M]

The time taken by the light reflected from sun to reach on earth will be 8.4 minutes.

To find the answer, we need to know about the distance travelled by light.

<h3>How to find the time taken by the light reflected from sun to reach on earth?</h3>

So, in order to solve this problem, we must first know how far the moon is from Earth and how far the Sun is from the moon.
These distances are given as 3.8×10^5 km (Earth-Moon) and 1.5×10^8 km (Sun- Earth).
Since the Moon and Sun are on opposite sides of Earth during a full moon, the light's distance traveled equals,

$d=(1.5*10^8km)+2(3.8*10^5km)=1.51*10^8km=1.51*10^{11}m$

As we know that light travels at a speed of 300,000 km per second. then, the time taken by the light reflected from sun to reach on earth will be,

$t=\frac{1.51*10^{11}}{3*10^8}=503.33 s\\t=\frac{503.33}{60}=8.4min$

Thus, the time it takes for the light from the Sun to reach Earth and be recognized as 8.4 minutes.

Learn more about distance here:

brainly.com/question/11495758

#SPJ4

6 0

2 years ago