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jarptica [38.1K]

3 years ago

5

A 1650 kg car accelerates at a rate of 4.0 m/s^2. How much force is the car's engine producing?

2 answers:

Westkost [7]3 years ago

7 0

F = M A

Force = (mass) x (acceleration)

= (1,650 kg) (4 m/s²) = 6,600 kg-m/s² = <em>6,600 Newtons</em>

Eduardwww [97]3 years ago

3 0

According to Newtons laws of motion F = MA. This means Force = Mass x Acceleration.

So...

4 x 1650 = 66,000
The units for Force is Newtons

So the answer is 66,000 Newtons

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A 3.50-meter length of wire with a cross-sectional area of 3.14 × 10-6 meter2 is at 20° Celsius. If the wire has a resistance of

maks197457 [2]

Answer:

$5.6\times 10^{-8}\ Ohm.m$

Explanation:

Resistivity is given by $\rho=\frac {AR}{L}$ where A is cross-sectional area, R is resistance, L is the length and $\rho$ is the reistivity. Substituting 0.0625 for R, 3.14 × 10-6 for A and 3.5 m for L then the resistivity is equivalent to

$\rho=\frac {3.14\times 10^{-6}\times 0.0625}{3.5}=5.60714285714285714285714285714285714285\times 10^{-8}\approx 5.6\times 10^{-8}\ Ohm.m$

8 0

3 years ago

A jet is circling an airport control tower at a distance of 20.6 km. An observer in the tower watches the jet cross in front of

lesya [120]

Answer:

197.76 m

Explanation:

r = Radius of the path = 20.6 km = $20.6\times 10^3\ m$

$\theta$ = The angle subtended by moon = $9.6\times 10^{-3}\ rad$

Distance traveled is given by

$s=r\times\theta$

$\Rightarrow s=20.6\times 10^3\times 9.6\times 10^{-3}$

$\Rightarrow s=197.76\ m$

The distance traveled by the jet is 197.76 m

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3 years ago

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3 years ago

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Some hypothetical alloy is composed of 12.5 wt% of metal A and 87.5 wt% of metal B. If the densities of metals A and B are 4.27

densk [106]

Answer:

The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.

Explanation:

Given that,

Weight of metal A = 12.5%

Weight of metal B = 87.5%

Length of unit cell = 0.395 nm

Density of A = 4.27 g/cm³

Density of B= 6.35 g/cm³

Weight of A = 61.4 g/mol

Weight of B = 125.7 g/mol

We need to calculate the density of the alloy

Using formula of density

$\rho=n\times\dfrac{m}{V_{c}\times N_{A}}$

$n=\dfrac{\rho\timesV_{c}\times N}{m}$ ....(I)

Where, n = number of atoms per unit cells

m = Mass of the alloy

V=Volume of the unit cell

N = Avogadro number

We calculate the density of alloy

$\rho=\dfrac{1}{\dfrac{12.5}{4.27}+\dfrac{87.5}{6.35}}\times100$

$\rho=5.98$

We calculate the mass of the alloy

$m=\dfrac{1}{\dfrac{12.5}{61.4}+\dfrac{87.5}{125.7}}\times100$

$m=111.15$

Put the value into the equation (I)

$n=\dfrac{5.9855\times(0.395\times10^{-9}\times10^{2})^3\times6.023\times10^{23}}{111.15}$

$n=1.99\approx 2\ atoms/cell$

Hence, The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.

5 0

3 years ago