All categories

2 years ago

4. The 50-kg crate shown in Fig. rests on a horizontal surface for which the coefficient of

Physics

1 answer:

laila [671]2 years ago

5 0

Answer:

5.057 m/s^2

Explanation:

Force of kinetic friction = .3 = F /normal force

.3 = F /(50* 9.81) F of friction = 147.15

Net force = 400 - 147.5 = 252.85 N

F = m * a

252.85 = 50 * a a = 5.057 m/s^2

You might be interested in

A jet is travelling at a speed of 1200 km/h and drops cargo from a height of 2.5 km above the ground Calculate the time it takes

OLEGan [10]

a) Time of flight: 22.6 s

To calculate the time it takes for the cargo to reach the ground, we just consider the vertical motion of the cargo.

The vertical position at time t is given by

$y(t) = h +u_y t - \frac{1}{2}gt^2$

where

h = 2.5 km = 2500 m is the initial height

$u_y = 0$ is the initial vertical velocity of the cargo

g = 9.8 m/s^2 is the acceleration of gravity

The cargo reaches the ground when

$y(t) = 0$

So substituting it into the equation and solving for t, we find the time of flight of the cargo:

$0 = h - \frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(2500)}{9.8}}=22.6 s$

b) 7.5 km

The range travelled by the cargo can be calculated by considering its horizontal motion only. In fact, the horizontal motion is a uniform motion, with constant velocity equal to the initial velocity of the jet:

$v_x = 1200 km/h \cdot \frac{1000 m/km}{3600 s/h}=333.3 m/s$

So the horizontal distance travelled is

$d=v_x t$

And if we substitute the time of flight,

t = 22.6 s

We find the range of the cargo:

$d=(333.3)(22.6)=7533 m = 7.5 km$

7 0

3 years ago

Please help me!

Kay [80]

Runoff because the mud is a liquid and moves on an amount of water that is in it like with quicksand

5 0

3 years ago

Read 2 more answers

Susan's 12.0 kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30∘ above the

Pavlova-9 [17]

Answer:1.71 m/s

Explanation:

Given

mass of Susan $m=12 kg$

Inclination $\theta =30^{\circ}$

Tension $T=29 N$

coefficient of Friction $\mu =0.18$

Resolving Forces Along x axis

$F_x=T\cos \theta -f_r$

where $f_r=friction\ Force$

$F_y=mg-N-T\sin \theta$

since there is no movement in Y direction therefore

$N=mg-T\sin \theta$

and $f_r=\mu N$

Thus $F_x=T\cos \theta -\mu N$

$F_x=29\cos (30)-\0.18\times (12\times 9.8-29\sin (30))$

$F_x=25.114-18.558$

$F_x=6.556 N$

Work done by applied Force is equal to change to kinetic Energy

$F_x\cdot x=\frac{1}{2}\cdot mv_f^2-\frac{1}{2}\cdot mv_i^2$

$6.556\times 2.7=\frac{1}{2}\cdot 12\times v_f^2$

$v_f^2=\frac{6.556\times 2.7\times 2}{12}$

$v_f^2=2.95$

$v_f=1.717 m/s$

8 0

3 years ago

An automobile rounds a curve of radius 50.0 m on a flat road.

bixtya [17]

Answer:

14m/s

Explanation:

Given parameters:

Radius of the curve = 50m

Centripetal acceleration = 3.92m/s²

Unknown:

Speed needed to keep the car on the curve = ?

Solution:

The centripetal acceleration is the inwardly directly acceleration needed to keep a body along a curved path.

It is given as;

a = $\frac{v^{2} }{r}$

a is the centripetal acceleration

v is the speed

r is the radius

Now insert the parameters and find v;

v² = ar

v² = 3.92 x 50 = 196

v = √196 = 14m/s

6 0

3 years ago

Calculate the acceleration of a bus that goes

Svet_ta [14]

The acceleration of the bus is 1.11 meters per second square to the direction of motion

Explanation:

Acceleration is the rate of change of velocity

The formula of the acceleration is $a=\frac{v_{2}-v_{1}}{t}$ , where

$v_{1}$ is the initial velocity

$v_{2}$ is the final velocity
t is the time

A bus that goes from 10 km/h to a speed of 50 km/h in 10 seconds

→ $v_{1}$ = 10 km/h

→ $v_{2}$ = 50 km/h

→ t = 10 seconds

Change the unite of the time from seconds to hour

→ 1 hour = 60 × 60 = 3600 seconds

→ 10 seconds = $\frac{10}{3600}=\frac{1}{360}$ hour

Substitute these values in the formula of the acceleration above

→ $a=\frac{50-10}{\frac{1}{360}}$

→ a = 14400 km/h²

To change the unit of acceleration to meter per second change the

kilometer to meter and the hour to seconds

→ 1 km = 1000 m

→ 1 hour = 3600 seconds

→ $a=\frac{14400*1000}{(3600)^{2}}=\frac{10}{9}$

→ a = 1.11 m/sec².

The acceleration of the bus is 1.11 meters per second square to the direction of motion

Learn more:

You can learn more about the acceleration in brainly.com/question/6323625

#LearnwithBrainly

4 0

3 years ago