A 34 kg block slides with an initial speed of 9 m/s up a
rampinclined at an angle of 10o with the horizontal.
Thecoefficient of kinetic friction between the block and the ramp
is0.6. Use energy conservation to find the distance the block
slidesbefore coming to rest.

Answer:

The right answer is (D) 340m

Explanation:

If the package is dropped from a helicopter moving upward and the air resistance is negligible, the only acting force in the package is the gravity force. Therefore you can consider this as a Vertical Projectile Motion problem.

The initial speed of the package V₀ is 15m/s.

The acceleration of the package G is 9.8m/s².

The initial hight is Y₀. Therefore:

Y(t)=Y₀+V₀·t-0.5·G·t²

But we know than T(10s)=0m

0m=Y₀+150m-490m

Y₀=340m

Answer:

Time taken by motorboat to reach [tex]17.0m/s[/tex] equals 13 seconds.

Explanation:

From the first equation of kinematics we have

[tex]v=u+at[/tex]

where,

'v' is the final speed of the accelerating object

'u' is the initial speed of the object

'a' is the accleration of the object

't' is the time for which the object accelerates

Applying the given values in the equation above we get

[tex]17=4+1.0\times t\\\\\\\therefore t=17-4=13seconds[/tex]

Answer:

Acceleration of the cart will be [tex]a=2m/sec^2[/tex]

Explanation:

We have given force F = 40 N

Mass of the cart = 20 kg

From newton's second law we know that force, mass and acceleration are related to each other

From second law of motion force on any object moving with acceleration a is given by

F = ma, here m is mass and a is acceleration

So [tex]40=20\times a[/tex]

[tex]a=2m/sec^2[/tex]

Answer:

The maximum height of the ball is 20 m. The ball needs 2 s to reach that height.

Explanation:

The equation that describes the height and velocity of the ball are the following:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the ball at time t

y0 = initial height

v0 = initial velocity

t = time

g = acceleration

v = velocity at time t

When the ball is at its maximum height, its velocity is 0, then, using the equation of the velocity, we can calculate the time at which the ball is at its max-height.

v = v0 + g · t

0 = 20 m/s - 9.8 m/s² · t

-20 m/s / -9.8 m/s² = t

t = 2.0 s

Then, the ball reaches its maximum height in 2 s.

Now,  we can calculate the max-height obtaining the position at time t = 2.0 s:

y = y0 + v0 · t + 1/2 · g · t²

y = 0 m + 20 m/s · 2 s - 1/2 · 9,8 m/s² · (2 s)²

y = 20 m

The maximum height reached by the ball is 20.4 meters, and it takes approximately 2.04 seconds to reach this height.

When a ball is thrown vertically into the air with an initial velocity of 20 m/s, we can calculate the maximum height using the kinematic equation:

[tex]v^2 = u^2 + 2gh,[/tex]

where v is the final velocity (0 m/s at the highest point), u is the initial velocity (20 m/s), g is the acceleration due to gravity (9.81 m/s2), and h is the maximum height. Solving for h gives us:

[tex]h = u^2 / (2g).[/tex]

By substituting the values we get:

[tex]h = (20 m/s)^2 / (2 * 9.81 m/s^2) = 20.4 m.[/tex]

To find the time needed to reach the maximum height, we use the equation:

v = u + gt,

Solving for t when v is 0 m/s, we get:

t = u / g = 20 m/s / 9.81 m/s2 = approx. 2.04 seconds.

Thus, the maximum height of the ball is 20.4 meters and the time needed to reach the maximum height is approximately 2.04 seconds.

Answer:

Focal length of the lens, f = - 68 mm

Explanation:

Given that,

Power of a lens, P = -14.50 D

We need to find the focal length of the lens. We know that the focal length and the power of lens has inverse relationship. Mathematically, it is given by :

[tex]f=\dfrac{1}{P}[/tex]

f is the focal length of the lens

[tex]f=\dfrac{1}{-14.50}[/tex]

f = -0.068 m

or

f = -68 mm

So, the focal length of the lens is (-68 mm). Hence, this is the required solution.

Answer:

The inicial height of the apple is 1.22 meters

Explanation:

Using the equation for conservarion of mechanical energy:

[tex]E=V+K=constant[/tex]

[tex]K_i=\frac{1}{2}mv_i^2[/tex] where v is the velocity

[tex]V=mgh[/tex]where h is the height

We equate the initial mechanical energy to the final:

Since [tex]v_0=0\ and h_f=0 [/tex]:

[tex]\frac{1}{2}mv_0^2+mgh_0= \frac{1}{2}mv_f^2+mgh_f\\gh_0= \frac{1}{2}v_f^2[/tex]

Solving for h:

[tex]h_0=\frac{4.9^2}{2g}= 1.22 m[/tex]

Answer:22.62 m/s

Explanation:

Given

two balls are separated by a distance of 50 m

Alex throws  the ball from a height of 50 m with a velocity of 8 m/s and Gary launches a ball with some velocity  exactly at the same time.

ball  from ground travels a distance of 25 m in t sec

For Person on tree  

[tex]25=ut+\frac{1}{2}gt^2[/tex]

[tex]25=8t+\frac{1}{2}\times 9.81\times t^2--------1[/tex]

For person at ground

[tex]23.5=ut-\frac{1}{2}gt^2---------2[/tex]

Solve equation (1)

[tex]50=16t+9.81t^2[/tex]

[tex]9.81t^2+16t-50=0[/tex]

[tex]t=\frac{-16\pm\sqrt{256+4\times 50\times 9.81}}{2\times 9.81}=\frac{47.1-16}{19.62}=1.58 s[/tex]

put the value of t in equation 2

[tex]23.5=u\times 1.58-\frac{9.81\times 1.58^2}{2}[/tex]

[tex]u=\frac{35.744}{1.58}=22.62 m/s[/tex]

Answer:

Explanation:

Force = mass x acceleration

[tex]\overrightarrow{F}=m\overrightarrow{a}[/tex]

Force is always vector and acceleration also vector but the mass is a saclar quanity.

here, the direction of force vector is same as the direction of acceleration vector but the magnitude of force depends on the magnitude of mass of the body.

Is mass is more, force is also more.

Thus, the mass is like an indicator of the magnitude of force.

Answer:

[tex]\rho = 2.39 g/mL[/tex]

[tex]\Delta \rho = 1.53 \times 10^{-3} mL[/tex]

Explanation:

As we know that density is the ratio of mass and volume of the object

here we know that

mass of the rock is

[tex]m = 750 g[/tex]

volume of the rock is given as

[tex]V = V_1 - V_o[/tex]

here we know that

[tex]V_1 = 813.2 \pm 0.1 mL[/tex]

[tex]V_2 = 500.0 \pm 0.1 mL[/tex]

now we have

[tex]V = 313.2 \pm 0.2 mL[/tex]

now density is given as

[tex]\rho = \frac{750}{313.2}[/tex]

[tex]\rho = 2.39 g/mL[/tex]

now uncertainty of density is given as

[tex]\Delta \rho = \frac{\Delta V}{V} \rho[/tex]

[tex]\Delta \rho = \frac{0.2}{313.2}(2.39)[/tex]

[tex]\Delta \rho = 1.53 \times 10^{-3} mL[/tex]

The density of the rock, given its mass is 750 g and the volume of water displaced is 313.2 mL, is 2.394 g/mL. The uncertainty in this measurement is ± 0.0015 g/mL, considering an uncertainty of ± 0.1 mL for both the initial and final volume measurements.

Given that the mass (m) of the rock is 750 g, the initial volume of water (V0) is 500.0 mL, and the volume of water plus the rock (V1) is 813.2 mL, we can determine the density (D) and the uncertainty in the density.

Using the formula:

D = m / (V1 - V0)

Therefore, D = 750 g / (813.2 mL - 500.0 mL)

= 750 g / 313.2 mL

= 2.394 g/mL.

Using the uncertainties in V0 and V1, which are both ± 0.1 mL. Since we subtract these volumes, the total volume uncertainty is ± (0.1 mL + 0.1 mL) = ± 0.2 mL. Thus, the uncertainty in the density (ΔD) can be approximated by the formula:

ΔD = D × (ΔV / (V1 - V0))

where ΔV is the total volume uncertainty. Substituting the values, we get ΔD = 2.394 g/mL × (0.2 mL / 313.2 mL) = ± 0.00153 g/mL (rounded to four significant figures).

Therefore, the estimated density of the rock is 2.394 ± 0.0015 g/mL.

Answer:

Football will be in air for 2.8139 sec

Explanation:

We have given maximum height h = 9.7 meters

Angle of projection [tex]\Theta =38^{\circ}[/tex]

We know that maximum height is given by [tex]h=\frac{u^2sin^2\Theta }{2g}[/tex]

So [tex]9.7=\frac{u^2sin^{2}38^{\circ} }{2\times 9.8}[/tex]

[tex]u^2=501.5842[/tex]

u = 22.396 m/sec

Time of flight is given by

[tex]T=\frac{2usin\Theta }{g}=\frac{2\times 22.396\times \times sin38^{\circ}}{9.8}=2.8139sec[/tex]

Answer:

6.86 cm

Explanation:

Given:

Assume:

When the first bead rests on the table, then electrostatic force due to the second bead acts on it in the upward direction, Normal force acts in the upward direction and its weight in the downward direction.

So, using Newton's second law on the first bead resting on the table, we have

[tex]F+N-W=0\\[/tex]

At the closest distance of the second bead to the first bead, it just lifts off the table and the normal force becomes zero.

[tex]\therefore F-W=0\\\Rightarrow F=W\\\Rightarrow \dfrac{kqQ}{r^2}=mg\\\Rightarrow r^2=\dfrac{kqQ}{mg}\\\Rightarrow r^2=\dfrac{9\times 10^9\times 2.9\times 10^{-9}\times 5.3\times 10^{-9}}{3\times 10^{-6}\times 9.8}\\\Rightarrow r^2=4.70\times 10^{-3}\\\textrm{Taking square root on both the sides}\\r = \pm 0.0686\ m\\\textrm{Since the distance is never negative}\\\therefore r = 0.0686\ m\\\Rightarrow r = 6.86\ cm[/tex]

Hence, the centers of the two beads must be brought closest to 6.86 cm before the lower bead is lifted off the table.

Answer:

T = 61.06 °C  

Explanation:

given data:

a thin metal bar consist of 5 different material.

thermal conductivity of ---

K {steel} = 16 Wm^{-1} k^{-1}

K brass = 125 Wm^{-1} k^{-1}

K copper = 401 Wm^{-1} k^{-1}

K aluminium =30Wm^{-1} k^{-1}

K silver = 427 Wm^{-1} k^{-1}

[tex]\frac{d\theta}{dt} = \frac{KA (T_2 -T_1)}{L}[/tex]

WE KNOW THAT

[tex]\frac{l}{KA} = thermal\ resistance[/tex]

total resistance of bar = R steel + R brass + R copper + R aluminium + R silver

[tex]R_{total} =\frac{1}[A} [\frac{0.02}{16} +\frac{0.03}{125} +\frac{0.01}{401} +\frac{0.05}{30} +\frac{0.01}{427}][/tex]

[tex]R_{total} =\frac{1}[A} * 0.00321[/tex]

let T is the temperature at steel/brass interference

[tex]\frac{d\theta}{dt}[/tex] will be constant throughtout the bar

therefore we have

[tex]\frac{100-0}{R_{total}} = \frac{100-T}{R_{steel}}[/tex]  

[tex]\frac{100-0}{0.00321} *A = \frac{100-T}{0.00125} *A[/tex]

solving for T  we get

T = 61.06 °C  

Answer:

11.306 nm/s

or

113.06 atomic layers/sec

Explanation:

Hello!

First we need to know how much an inch equals in nanometers and a day in seconds:

Since 1inch = 2.54cm and 1cm=10^7nm

     1 inch = 2.54 * 10^7 nm

Also 1day = 24hours = 24*60minutes = 24*60*60seconds

   1 day = 86.4 * 10^3 s

Therefore the rate at which the hair grows in nanometers per seconds is:

    1/26 in/day = (1/26) * (2.54*10^7)/(86.4*10^3) = 11.306 nm/s

Now, if 1 atomic layer = 0.1 nm this means that 1 nm = 10 atomic layers.

Therefore:

The rate in atomic layers is

11.306 nm/s = 11.306 (10 atomic layers)/s = 113.06 atomic layers/sec

Answer:

absolute pressure =  1.07 lbft/in^2

Explanation:

given data:

vaccum gauge reading h = 27.86 inch = 2.32 ft

we know that

gauge pressure p is given as

[tex]p = \rho gh[/tex]

[tex]p = 848 \ lb/ft^3 * 32 \ ft/s^2 *2.32 ft  = 62955.52 \ lbft/s^2 * 1/ft^2[/tex]

we know that  [tex]1\  lb ft\s^2 = \frac{1}{32.174}\  lbft[/tex]

1 ft = 12 inch

therefore [tex]p = 62955.52 * \frac{1}{32.174} * \frac{1}{12^2}\ lbf/in^2[/tex]

               [tex]p = 13.59\  lbft/in^2[/tex]

[tex]P_{atm} = 14.66\  lbf/in^2[/tex]

so absolute pressure = [tex]P_{atm} - p[/tex]

                                   = 14.66 - 13.59 = 1.07 lbft/in^2

Answer:

1.05 ms⁻²

Explanation:

Acceleration = change in velocity / Time

Change in velocity = Final velocity - initial velocity

= 1.77 - (-1.29)

= 1.77 + 1.29

= 3.06 m/s

Time = 2.91

Acceleration = 3.06 / 2.91

= 1.05 ms⁻² .

Answer:

The correct option is 'D': 9

Explanation:

We know that the magnitude of the centripetal acceleration of  a body moving in circular orbit of radius 'r' with speed 'v' is given by

[tex]a_{c}=\frac{v^{2}}{r}[/tex]

Now when the speed of the body is tripled the speed becomes [tex]3v[/tex]

Hence the new centripetal acceleration is obtained as

[tex]a'_{c}=\frac{(3v)^{2}}{r}\\\\a'_{c}=\frac{9v^{2}}{r}=9a_{c}[/tex]

Thus we can see that the new centripetal acceleration becomes 9 times the oroginal value.

The acceleration of a body traveling in a circular route is known as centripetal acceleration. The magnitude of the centripetal acceleration increases by a factor of 9.

The acceleration of a body traveling in a circular route is known as centripetal acceleration. Because velocity is a vector quantity. It has both a magnitude and a direction.

When a body moves on a circular route, its direction changes constantly, causing its velocity to vary, resulting in acceleration.

Mathematically it is given as,

[tex]\rma_c=\frac{v^2}{r} \\\\ a_c'=\frac{(3v)^2}{r} \\\\ \rm v=9\frac{v^2}{r}\\\\ a_c'=9a_c[/tex]

Hence the magnitude of the centripetal acceleration increases by a factor of 9. Option c is correct.

To learn more about centripetal acceleration refer to the link;

https://brainly.com/question/17689540

Answer:

a) Please, see the attched figure

b) Total displacement R = (78.3 km; -4.8 km)

c) R = (78.4 km * cos (-3.5°); 78.4 km * sin (-3.5°))

d) The hippo is 78.4 km from his starting point.

The total distance traveled is 102 km

Explanation:

a)Please, see the attached figure.

b) The vector A can be expressed as:

A = (magnitude * cos α; magnitude * sin α)

Where

magnitude = 42 km

α= 0

Then,

A = (42 km ; 0) or 42 km i

In the same way, we can proceed with the other vectors:

B = ( Bx ; By)

where

(apply trigonometry of right triangles: sen α = opposite / hypotenuse and

cos α = adjacent / hypotenuse. See the figure to determine which component of vector B is the opposite and adjacent side to α)

Bx = 28 km * sin 25 = 11.8 km

By = 28 km * cos 25 = -25.4 km (it has to be negative since it is directed towards the negative vertical region according to our reference system)

B = (11.8 km; -25.4 km) or 11.8 km i - 25.4 km j

C = (Cx; Cy)

where

Cx = 32 km * cos 40° = 24.5 km

Cy = 32 km * sin 40 = 20.6 km

C = (24.5 km; 20.6 km)

Then:

R = A+B+C = (42 km + 11.8 km + 24.5 km; 0 - 25.4 km + 20.6 km)

= (78.3 km; -4.8 km) or 78.3 km i -4.8 km j

c) R = (78.3 km; -4.8 km)

The magnitude of R is:

[tex]magnitude = \sqrt{(78.3)^{2 }+ (-4.8)^{2}}= 78.4 km[/tex]

Using trigonometry, we can calculate the angle:

Knowing that

tan α = opposite / adjacent

and that

opposite = Ry = -4.8 km

adjacent = Rx = 78.3 km

Then:

tan α = -4.8 km / 78.4 km

α = -3.5°

We can now write the vector R in magnitude and direction form:

R = (78.4 km * cos (-3.5°); 78.4 km * sin (-3.5°))

d) The displacement of the hipo relative to the starting point is the magnitude of vector R calculated in c):

magnitude R = 78. 4 km

The total distance traveled is the sum of the magnitudes of each vector:

Total distance = 42 km + 28 km + 32 km = 102 km  

Answer:

(C) 3.78 seconds

Explanation:

At the highest point, the velocity is equal to 0m/s

[tex]v_{f}=v_{o}-gt[/tex]

[tex]t=\frac{v_{o}}{g}[/tex]  ; t is the time to reach the highest point

The  time the ball takes to return to its starting point after the ball  reach its maximum height is the same:

[tex]T_{descent}=t=\frac{v_{o}}{g}=\frac{37}{9.81}=3.78s[/tex]

Answer:

c) Both a) and b) are the combinations that have a negative velocity.

Explanation:

The velocity is given by this equation:

v = Δx / Δt

Where

Δx = final position - initial position

Δt = elapsed time

Now let´s evaluate the options. We have to find those combinations in which  Δx < 0 since Δt is always positive.

a) If the initial position, x, is positive and you move towards the origin, the final position will be a smaller value than x. Then:

                          final position < initial position

final position - initial position < 0 The velocity will be negative.

b) If x is negative and you move away from the origin, the final position will be a more negative number than x. Again:

                          final position < initial position

final position - initial position < 0 The velocity will be negative.

Let´s do an example to show it:

initial position = -5

final position = -10 (since you moved away from the origin)

final position - initial position = -10 -(-5) = -5

d) If x is positive and you move away from the origin, the final position will be a greater value than the initial position. Then:

                          final position > initial position

final position - initial position > 0 The velocity will be positive.

e) If x is negative and you move towards the origin, the final position will be a greater value than the initial position. Then:

                          final position > initial position

final position - initial position > 0 The velocity will be positive.

Let´s do an example:

initial position = -10

final position = -5

final postion - initial position = -5 - (-10) = 5

or with final position = 0

final postion - initial position = 0 -(-10) = 10

And so on.

The right answer is c) Both a) and b) are the combinations that have a negative velocity.

(a) 18.9 m/s

The motion of the stone consists of two independent motions:

- A horizontal motion at constant speed

- A vertical motion with constant acceleration ([tex]g=9.8 m/s^2[/tex]) downward

We can calculate the components of the initial velocity of the stone as it is launched from the ground:

[tex]u_x = v_0 cos \theta = (25.0)(cos 41.0^{\circ})=18.9 m/s\\u_y = v_0 sin \theta = (25.0)(sin 41.0^{\circ})=16.4 m/s[/tex]

The horizontal velocity remains constant, while the vertical velocity changes due to the acceleration along the vertical direction.

When the stone reaches the top of its parabolic path, the vertical velocity has became zero (because it is changing direction): so the speed of the stone is simply equal to the horizontal velocity, therefore

[tex]v=18.9 m/s[/tex]

(b) 22.2 m/s

We can solve this part by analyzing the vertical motion only first. In fact, the vertical velocity at any height h during the motion is given by

[tex]v_y^2 - u_y^2 = 2ah[/tex] (1)

where

[tex]u_y = 16.4 m/s[/tex] is the initial vertical velocity

[tex]v_y[/tex] is the vertical velocity at height h

[tex]a=g=-9.8 m/s^2[/tex] is the acceleration due to gravity (negative because it is downward)

At the top of the parabolic path, [tex]v_y = 0[/tex], so we can use the equation to find the maximum height

[tex]h_{max} = \frac{-u_y^2}{2a}=\frac{-(16.4)^2}{2(-9.8)}=13.7 m[/tex]

So, at half of the maximum height,

[tex]h = \frac{13.7}{2}=6.9 m[/tex]

And so we can use again eq(1) to find the vertical velocity at h = 6.9 m:

[tex]v_y = \sqrt{u_y^2 + 2ah}=\sqrt{(16.4)^2+2(-9.8)(6.9)}=11.6 m/s[/tex]

And so, the speed of the stone at half of the maximum height is

[tex]v=\sqrt{v_x^2+v_y^2}=\sqrt{18.9^2+11.6^2}=22.2 m/s[/tex]

(c) 17.4% faster

We said that the speed at the top of the trajectory (part a) is

[tex]v_1 = 18.9 m/s[/tex]

while the speed at half of the maximum height (part b) is

[tex]v_2 = 22.2 m/s[/tex]

So the difference is

[tex]\Delta v = v_2 - v_2 = 22.2 - 18.9 = 3.3 m/s[/tex]

And so, in percentage,

[tex]\frac{\Delta v}{v_1} \cdot 100 = \frac{3.3}{18.9}\cdot 100=17.4\%[/tex]

So, the stone in part (b) is moving 17.4% faster than in part (a).

Answer:

Clockwise, converging towards

Explanation:

At the center of surface , system with low pressure in Northern hemisphere have air around the center that rotates in anti clockwise direction.

As a result of low pressure, the air is directed slightly inwards thus converges towards the center of the system.

In the high pressure systems, air rotates in clockwise direction and diverges away from the center of the system.

Answer:

a) 30.84m/s

b) 348.32Hz

c) 32.34m/s

d) 289.69Hz

Explanation:

a) If 1 mile=1609,34m, and 1 hour=3600 seconds, then 69mph=69*1609.34m/3600s=30.84m/s

b) Based on Doppler effect:

/*I will take as positive direction the vector [tex]\vec r_{observer}-\vec r_{emiter}[/tex] */

[tex]f_{observed}=(\frac{v_{sound}-v_{observed}}{v_{sound}-v_{emited}})f_{emited}[/tex]

[tex]f_{observed}=(\frac{343m/s-0m/s}{343m/s-30.84m/s})317Hz=348.32Hz[/tex]

c) [tex]350Hz=(\frac{343m/s-0m/s}{343m/s-v_{ambulance}})317Hz, V_{ambulance}=343m/s-\frac{317Hz}{350Hz}.343m/s=32.34m/s[/tex]

d) [tex]f_{observed}=(\frac{343m/s-0m/s}{343m/s+32.34m/s})317Hz=289.69Hz[/tex]

Answer:

(a). The car's total displacement from its starting point is 76.58 m.

(b). The angle of the car's total displacement measured from its starting direction is 36.81°.

Explanation:

Given that,

Distance = 47 km in east

Distance = 23 km in north

Angle = 32° east of north

Distance = 27 km

According to figure,

Angle = 90-32 = 58°

(a). We need to calculate the magnitude of the car's total displacement from its starting point

Using Pythagorean theorem

[tex]AC=\sqrt{AB^2+BC^2}[/tex]

[tex]AC=\sqrt{(47+27\cos58)^2+(23+27\sin58)^2}[/tex]

[tex]AC=76.58\ m[/tex]

The magnitude of the car's total displacement from its starting point is 76.58 m.

(b). We need to calculate the angle (from east) of the car's total displacement measured from its starting direction

Using formula of angle

[tex]\tan\theta=\dfrac{y}{x}[/tex]

put the value into the formula

[tex]\theta=tan^{-1}\dfrac{23+27\sin58}{47+27\cos58}[/tex]

[tex]\theta=tan^{-1}0.7486[/tex]

[tex]\theta=36.81^{\circ}[/tex]

Hence, (a). The car's total displacement from its starting point is 76.58 m.

(b). The angle of the car's total displacement measured from its starting direction is 36.81°.

Answer:

a) Acceleration of runner is 1.33 m/s²

b)  Acceleration of motorcycle is 2.85 m/s²

c) The motorcycle moves 84.21-2.94 = 81.06 m farther than the runner.

Explanation:

t = Time taken

u = Initial velocity = 0

v = Final velocity

s = Displacement

a = Acceleration

Equation of motion

[tex]v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{2.8-0}{2.1}\\\Rightarrow a=1.33\ m/s^2[/tex]

Acceleration of runner is 1.33 m/s²

[tex]v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{43-37}{2.1}\\\Rightarrow a=2.85\ m/s^2[/tex]

Acceleration of motorcycle is 2.85 m/s²

[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{2.8^2-0^2}{2\times 1.33}\\\Rightarrow s=2.94\ m[/tex]

The runner moves 2.94 m

[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{43^2-37^2}{2\times 2.85}\\\Rightarrow s=84.21\ m[/tex]

The motorcycle moves 84.21 m

The motorcycle moves 84.21-2.94 = 81.06 m farther than the runner.

Answer:

Not the right answer in the options, speed is 4.47 m/s, and the procedure is coherent with option A

Explanation:

Answer A uses mass and velocity units, which are momentum units. By using the conservation of momentum:

.[tex]p_{initial} =p_{final} \\m_{Tom}*v_{Tom}+m_{raft}*v_{raft}=(m_{Tom}+m_{raft})*v_{both} \\70*10+130*1.5 kg*m/s=895kg*m/s\\v_{both}=\frac{895 kg*m/s}{200 kg} =4.47 m/s[/tex]

Since Tom stays in the raft, then both are moving with the same speed. From the options, the momentum is in agreement with option A, however, the question asks for speed.

Final answer:

The work done by the man in pushing the lawn mower is 699.245 J, calculated by determining the horizontal force component and multiplying by the distance pushed.

Explanation:

To calculate how much work is done by the man in pushing the lawn mower, we need to consider only the component of the force that acts in the direction of the movement. Since 37% of the 195 N force is directed downward, only the remaining 63% is contributing to the horizontal movement. Therefore, the horizontal component of the force is 0.63 × 195 N = 122.85 N.

The formula to calculate work (W) is W = force (F) × distance (d) × cosine(θ), where θ is the angle between the force and the direction of movement. In this case, the force and movement are in the same direction, so θ = 0 and cosine(θ) = 1. Thus, the work done is:

W = 122.85 N × 5.7 m × 1 = 699.245 J

In terms of energy expended while pushing a lawn mower, this work is a relatively small amount when compared to a person's daily intake of food energy.

Answer:

1.43 x 10¹⁷.

They will accelerate away from each other.

Explanation:

Force on each charged sphere F = mass x acceleration

= 8.55 x 10⁻³ x 25 x 9.8

= 2.095 N

Let Q be the charge on each sphere

F = [tex]\frac{9\times10^9\times Q^2}{(15\times10^{-2})^2}[/tex]

2.095 =[tex]\frac{9\times10^9\times Q^2}{(15\times10^{-2})^2}[/tex]

Q² =[tex]\frac{2.095\times(15)^2\times10^{-4}}{9\times10^9}[/tex]

Q = 2.289 X 10⁻⁶

No of electrons = Charge / charge on a single electron

= [tex]\frac{2.289\times10^{-6}}{1.6\times10^{-19}}[/tex]

=1.43 x 10¹³.

They will accelerate away from each other.

The number of electrons would have to add to each sphere to accelerate at 25.0g is 1.43×10¹³ and accelerate away from each other.

Electric force is the force of attraction of repulsion between two bodies.

According to the Coulombs law, the force of attraction of repulsion between charged two bodies is directly proportional to the product of charges of them and inversely proportional to the square of distance between them. It can be given as,

[tex]F=\dfrac{KQ_1Q_2}{r^2}[/tex]

Here, (k) is the coulombs constant,  (q1 and q2) is the charges of two bodies and (r) is the distance between the two charges.

Two very small 8.55-g spheres, 15.0 cm apart from center to center, are charged by adding equal numbers of electrons to each of them.

Two spheres accelerate at 25.0g when released, and the mass of each sphere is 8.55 g. Thus, the force on the sphere can be given as,

[tex]F=8.55\times{10^{-3}}\times25\times9.8\\F=2.095\rm \;N[/tex]

Two very small spheres are 15.0 cm apart from center to center, are charged by adding equal numbers of electrons to each of them.As the charge on both the sphere is same (say Q) Put these values in the above formula as,

[tex]2.095=\dfrac{(9\times10^{9})QQ}{(15\times10^{-2})^2}\\Q=2.289\times10^{-6}\rm \;C[/tex]

It is known that the charge on one electron is 1.6×10⁻¹⁹ C. Thus the number of electron in the above charge is,

[tex]n=\dfrac{2.289\times10^{-6}}{1.6\times10^{-19}}\\n=1.43\times10^{13}[/tex]

The direction of acceleration of both the sphere will be opposite to each other. Thus, they accelerate away from each other.

Hence, the number of electrons would have to add to each sphere to accelerate at 25.0g is 1.43×10¹³ and accelerate away from each other.

Learn more about the electric force here;

https://brainly.com/question/14372859

Answer:

a) 4.45 m/s

b) 0.9 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

[tex]v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow u=\sqrt{v^2-2as}\\\Rightarrow u=\sqrt{0^2-2\times -9.81\times 1}\\\Rightarrow u=4.45\ m/s[/tex]

a) The vertical speed when the player leaves the ground is 4.45 m/s

[tex]v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-4.45}{-9.81}\\\Rightarrow t=0.45\ s[/tex]

Time taken to reach the maximum height is 0.45 seconds

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 1=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1\times 2}{9.81}}\\\Rightarrow t=0.45\ s[/tex]

Time taken to reach the ground from the maximum height is 0.45 seconds

b) Time the player stayed in the air is 0.45+0.45 = 0.9 seconds

Answer:

3.41 s

114 m

Explanation:

The object is falling in free fall, accelerated by the surface gravity of Earth. We can use the equation for position under constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

We set up a frame of reference with the origin at the point the object was released and the X axis pointing down. Then X0 = 0. Since the problem doesnt mention an initial speed we assume V0 = 0.

It travels 0.5h in the last 1 second of the fall. This means it also traveled in the rest of the time of the fall. t = t1 is the moment when it traveled 0.5*h.

0.5*h = 1/2 * a * t1^2

h = a * t1^2

It travels 0.5*h in 1 second.

h = X(t1 + 1) = 1/2 * a * (t1+1)^2

Equating both equations:

a * t1^2 = 1/2 * a * (t1+1)^2

We simplify a and expand the square

t1^2 = 1/2 * (t1^2 + 2*t1 + 1)

t1^2 - 1/2 * t1^2 - t1 - 1/2 = 0

1/2 * t1^2 - t1 - 1/2 = 0

Solving electronically:

t1 = 2.41 s

total time = t1 + 1 = 3.41.

Now

h = a * t1^2

h = 9.81 * 3.41^2 = 114 m

Answer:

[tex]Q=81.56\ W/m^2[/tex]

Explanation:

Given that

[tex]T_1= 210 K[/tex]

[tex]T_2= 150 K[/tex]

Emissivity of surfaces(∈) = 1

We know that heat transfer between two surfaces due to radiation ,when both surfaces are black bodies

[tex]Q=\sigma (T_1^4-T_2^4)\ W/m^2[/tex]

So now by putting the values

[tex]Q=\sigma (T_1^4-T_2^4)\ W/m^2[/tex]

[tex]Q=5.67\times 10^{-8}(210^4-150^4)\ W/m^2[/tex]

[tex]Q=81.56\ W/m^2[/tex]

So rate of heat transfer per unit area

[tex]Q=81.56\ W/m^2[/tex]