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3 years ago

A car has a mass of 1500 kg. If the force acting on the car is 6750 N to the east, what is the car’s acceleration?

Physics

1 answer:

Paha777 [63]3 years ago

8 0

The car's acceleration is equal to $4.5\frac{m}{s^{2}}$ to the east.

Why?

To calculate the car's acceleration, we need to use the given information about its mass and the force applied on it. We can use the following formula:

$Force=mass*acceleration\\\\6570N=1500Kg*acceleration\\\\acceleration=\frac{6750N}{1500Kg}=4.5\frac{m}{s^{2}}$

Hence, we have that the car's acceleration is equal to $4.5\frac{m}{s^{2}}$ to the east.

Have a nice day!

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A car passes point “A” and then 120 meters later. It’s velocity was measured 21 m/s. If it’s acceleration was constant at 0.853

Norma-Jean [14]

Recall that

${v_f}^2-{v_i}^2=2a\Delta x$

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So we have

$\left(21\dfrac{\rm m}{\rm s}\right)^2-{v_i}^2=2\left(0.853\dfrac{\rm m}{\mathrm s^2}\right)(120\,\mathrm m)$

$\implies v_i\approx\boxed{15.4\dfrac{\rm m}{\rm s}}$

(Normally, this equation has two solutions, but we omit the negative one because the car is moving in one direction.)

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3 years ago

A 265 g mass attached to a horizontal spring oscillates at a frequency of 3.40 Hz . At t =0s, the mass is at x= 6.20 cm and has

lara [203]

Answer:

The phase constant is 7.25 degree

Explanation:

given data

mass = 265 g

frequency = 3.40 Hz

time t = 0 s

x = 6.20 cm

vx = - 35.0 cm/s

solution

as phase constant is express as

y = A cosФ ..............1

here A is amplitude that is = $\sqrt{(\frac{v_x}{\omega })^2+y^2 }$ = $\sqrt{(\frac{35}{2\times \pi \times y})^2+6.2^2 }$ = 6.25 cm

put value in equation 1

6.20 = 6.25 cosФ

cosФ = 0.992

Ф = 7.25 degree

so the phase constant is 7.25 degree

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3 years ago

a basketball is tossed upwards with a speed of 5.0 m/s What is the maximum height reached by the basketball from its release poi

den301095 [7]

Answer:

1.275 m

Explanation:

Let the maximum height reached be h.

Here initial velocity, u = 5 m/s

Final velocity, V = 0

Use third equation of motion

V^2 = u^2 + 2 g h

0 = 25 - 2 × 9.8 × h

h = 25/19.6 = 1.275 m

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3 years ago

Read 2 more answers

A person of mass m is standing on the surface of the Earth, of mass M E . What is the acceleration that the Earth experiences du

Lana71 [14]

Answer:

$a_E=\dfrac{Gm}{r^2}$

Explanation:

$M_E$ = Mass of the Earth = 5.972 × 10²⁴ kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Earth = 6371000 m

m = Mass of person

The force on the person will balance the gravitational force

$M_Ea_E=\dfrac{GmM_E}{r^2}\\\Rightarrow a_E=\dfrac{Gm}{r^2}$

The acceleration that the Earth will feel is $a_E=\dfrac{Gm}{r^2}$

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3 years ago

A baseball is thrown with an initial velocity of 45.4 m/s at an angle of 31.2 ∘ .

sveticcg [70]

Answer:

V (initial vertical velocity) = 45.4 sin 31.2 = 23.52 m/s

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h = V^2 / (2 g) = 23.52^2 / 19.6 = 28.2 m max height

Check:

t = 28.2 / 9.8 = 2.88 sec time to reach max height

h = 23.52 * 2.88 - 1/2 g 2.88^2 = 27.1 m

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2 years ago