How much work must be done on a 750 kg car to slow it down from 1.0 x 10 km/h to 2 50.0 km/h?

Elden [556K]

A person who sells moongphali (groundnuts).

6 0

3 years ago

Henrietta is jogging on the side-walk at 3.05 m/s on the way to her physics class. Bruce realizes that she forgot her bag of bag

shusha [124]

Answer:

12.9121148614 m/s

35.9393048982 m

Explanation:

t = Time taken

u = Initial velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² = a

$s=ut+\frac{1}{2}at^2\\\Rightarrow 38=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{38\times 2}{9.81}}\\\Rightarrow t=2.78337865516\ s$

Time taken for the bag to fall is 2.78337865516 seconds

Time she has been jogging for

9+2.78337865516 = 11.78337865516 seconds

Total distance traveled by her

$s=vt\\\Rightarrow s=3.05\times 11.78337865516=35.9393048982\ m$

Henrietta is 35.9393048982 m away

Velocity of throwing

$\dfrac{35.9393048982}{2.78337865516}=12.9121148614\ m/s$

The velocity of throwing is 12.9121148614 m/s

3 0

4 years ago

How would you determine how much error there is between a vector addition and the real results

chubhunter [2.5K]

Desired operation: A + B = C; {A,B,C) are vector quantities.

Issue: {A,B} contain error (measurement or otherwise) 

Objective: estimate the error in the vector sum. 

Let A = u + du; where u is the nominal value of A and du is the error in A 
Let B = v + dv; where v is the nominal value of B and dv is the error in B 
Let C = w + dw; where w is the nominal value of C and dw is the error in C [the objective] 

C = A + B 

w + dw = (u + du) + (v + dv) 

w + dw = (u + v) + (du + dv) 

w = u+v; dw = du + dv 

The error associated with w is the vector sum of the errors associated with the measured quantities (u,v)

6 0

3 years ago

The electric flux through a square-shaped area of side 5 cm near a very large, thin, uniformly-charged sheet is found to be 3\ti

deff fn [24]

Answer:

Explanation:

Given

side of square shape $a=5\ cm$

Electric flux $\phi =3\times 10^{-5}\ N.m^2/C$

Permittivity of free space $\epsilon_0=8.85\times 10^{-12} \frac{C^2}{N.m^2}$

Flux is given by

$\phi =EA\cos \theta$

where E=electric field strength

A=area

$\theta$ =Angle between Electric field and area vector

$E=\frac{\phi }{A\cos (0)}$

$E=\frac{3\times 10^{-5}}{25\times 10^{-4}\times \cos(0)}$

$E=0.012\ N/C$

and Electric field by a uniformly charged sheet is given by

$E=\frac{\sigma }{2\epsilon_0}$

where $\sigma$ =charge density

$=\frac{\sigma }{\epsilon_0}$

$\sigma =0.012\times 8.85\times 10^{-12}$

$\sigma =2.12\times 10^{-13}\ C/m^2$

5 0

3 years ago

A friend pushes a sled across horizontal snow and when it gets up to speed the friend jumps on. After the friend jumps on, the s

Shalnov [3]

Answer:

v’ =( $\frac{1}{1+ \frac{M}{m} }$ ) v

we see that the greater the difference, the more the sled slows down.

friction force

Explanation:

When the man pushes the sled he does work and the sled acquires a speed and as long as it is supplied with an energy equal to the work of the chipping force with the snow, the speed is maintained.

When he jumps on the sled, a collision occurs and the initial moment

p₀ = mv

is increased by the increase in mass

m_f= (m + M_{man} ) v '

In this case there is no longer any external force applied and the only external force is friction, which causes the sled to stop, even when it is small, but the significant reduction in speed is due to the increase in masses.

p₀ = p_f

mv = (m + M_{man}) v '

v ’= $\frac{m}{m+M}$ v

v’ =( $\frac{1}{1+ \frac{M}{m} }$ ) v

Therefore, we see that the greater the difference, the more the sled slows down.

The only forces that act on the sled with the man are the friction that is responsible for the decrease in speed and weight with the normal

4 0

3 years ago