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zmey [24]
3 years ago
7

The engine of a 550kg car is producing a net force of 11,000N. What is its acceleration?

Physics
2 answers:
Fudgin [204]3 years ago
8 0

Answer:

20

Explanation:

(correct me if im wrong)

F = m×a

11000 = 550 × a

11000/550 = a

Acceleration =20

finlep [7]3 years ago
7 0

Answer:

Yeda

Explanation:

yshyei5rexwu drift hai used thakare I exam rest watu testing reddy

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A 1200 Kg car rounds a corner of radius r = 45m. If the coefficient of static friction between the ties and the road is us = 0.8
Vaselesa [24]

Answer:

The greatest speed of the car is 19.36m/s

Explanation:

The maximum speed the car will attain without skidding is given by:

F= uN = umg ...eq1

But F = mv^2/r

mv^2/r = umg

Dividing both sides by m, leaves you with:

V= Sqrt(ugr)

Where u = coefficient of static friction

g = acceleration due to gravity

r = raduis

Given:

U = 0.82

r=0.82

g= 9.8m/s

V = Sqrt(0.82 × 9.8 × 45)

V = Sqrt(374.85)

V = 19.36m/s

5 0
4 years ago
A system consists of three particles with these masses and velocities: mass 3.00 kg, moving north at 3.00 m/s; mass 4.00 kg, mov
lions [1.4K]

Answer:

the total momentum is 8 .2 kg m/s in north direction.

Explanation:

given,

mass(m₁) 3.00 kg, moving north at v₁ = 3.00 m/s

mass(m₂) 4.00 kg, moving south at v₂ =  3.70 m/s

mass(m₃) 7.00 kg, moving north at v₃ = 2.00 m/s

north as the positive axis

south as the negative axis

now

total momentum = m₁v₁ + m₂ v₂ + m₃ v₃

total momentum = 3 x 3 - 4 x 3.7 + 7 x 2

                           = 9 - 14.8 + 14

                           = 8 .2 kg m/s

hence, the total momentum is 8 .2 kg m/s in north direction.

7 0
4 years ago
A current of 12 amps is measured in a circuit with a total resistance of 9.0 ohms. What is the size of the voltage source that s
Serjik [45]
Data:
i (current) = 12 A
R (resistance) = 9.0 Ω
V (voltage) = ? (volts)

Formula:
V = R*i

Solving:
V = R*i
V = 9.0*12
\boxed{\boxed{V = 108\:volts}}\end{array}}\qquad\quad\checkmark
6 0
3 years ago
Estimate the electric field at a point 2.40 cm perpendicular to the midpoint of a uniformly charged 2.00-m-long thin wire carryi
nadya68 [22]

Answer:

E = 1.85*10^{12}\frac{N}{C}

Explanation:

Hi!

The perpendicular distance 2.4cm, is much less than the distance to both endpoints of the wire, which is aprox 1m. Then the edge effect is negligible at this field point, and we can aproximate the wire as infinitely long.

The electric filed of an infinitely long wire is easy to calculate. Let's call z the axis along the wire. Because of its simmetry (translational and rotational), the electric field E must point in the radial direction,  and it cannot depende on coordinate z. To calculate the field Gauss law is used, as seen in the image, with a cylindrical gaussian surface. The result is:

E = \frac{\lambda}{2\pi \epsilon_0 r}\\\lambda=\text{charge per unit length}=\frac{4.95 \mu C}{2 m} = 2.475 \frac{C}{m}\\r=\text{perpendicular distance to wire}\\\epsilon_0=8.85*10^{-12}\frac{C^2}{Nm^2}

Then the electric field at the point of interest is estimated as:

E = \frac{\22.475}{2\pi*( 8.85*10^{-12})*(2.4*10^{-2})}\frac{N}{C}=1.85*10^{12}\frac{N}{C}

6 0
4 years ago
When the cart moves on an incline at constant speed, it is in equilibrium; i.e., the net force on it is zero. Does it require mo
Amanda [17]

Answer:

It requires more tension to pull up the track

Explanation:

Net force must be zero to maintain constant velocity.

Weight force will always be pointed down the slope. Call it W

Friction force (Call it Ff) will be down slope when movement is up slope.

Friction force will be up slope when movement is down slope.

W and Ff are always positive numbers

Call the pulling force T

If Up slope is considered the positive direction

Moving up slope

Tu - Ff - W = 0

Tu = Ff + W

Moving down slope

Td + W - Ff = 0

Td = Ff - W

Ff + W > Ff - W therefore Tu > Td

5 0
3 years ago
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