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3 years ago

Ann is driving down a street at 61 km/h.

Physics

1 answer:

Wittaler [7]3 years ago

4 0

Answer:

12.267 seconds approximately.

Explanation:

The units can be simplified into m/s, in which case you would have 61000/3600. Simplify that to 16 and 17/18. This is your meters per second, so multiply that by .724 to get the answer.

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I need help with #40 need answer and show me what you multiplied please

ale4655 [162]

(0.5)×(0squared)×(3)=(1.5j)

3 0

3 years ago

A 0.29 kg particle moves in an xy plane according to x(t) = - 19 + 1 t - 3 t3 and y(t) = 20 + 7 t - 9 t2, with x and y in meters

Artist 52 [7]

Answer:

Part a)

$F = 7.76 N$

Part b)

$\theta = -137.7 degree$

Part c)

$\theta = -127.7 degree$

Explanation:

As we know that acceleration is rate of change in velocity of the object

So here we know that

$x = -19 + t - 3t^3$

$y = 20 + 7t - 9t^2$

Part a)

differentiate x and y two times with respect to time to find the acceleration

$a_x = \frac{d^2}{dt^2}(-19 + t - 3t^3)$

$a_x = \frac{d}{dt}(0 +1 - 9t^2)$

$a_x = -18t$

$a_y = \frac{d^2}{dt^2}(20 + 7t - 9t^2)$

$a_y = \frac{d}{dt}(0 +7 - 18t)$

$a_y = -18$

Now the acceleration of the object is given as

$\vec a = (-18t)\hat i + (-18)\hat j$

at t= 1.1 s we have

$\vec a = -19.8 \hat i - 18 \hat j$

now the net force of the object is given as

$\vec F = m\vec a$

$\vec F = (0.29 kg)(-19.8 \hat i - 18 \hat j)$

$\vec F = -5.74 \hat i - 5.22 \hat j$

now magnitude of the force will be

$F = \sqrt{5.74^2 + 5.22^2} = 7.76 N$

Part b)

Direction of the force is given as

$tan\theta = \frac{F_y}{F_x}$

$tan\theta = \frac{-5.22}{-5.74}$

$\theta = -137.7 degree$

Part c)

For velocity of the particle we have

$v_x = \frac{dx}[dt}$

$v_x = (0 +1 - 9t^2)$

$v_y = \frac{dy}{dt}$

$v_y = (0 +7 - 18t)$

now at t = 1.1 s

$\vec v = -9.89\hat i - 12.8 \hat j$

now the direction of the velocity is given as

$\theta = tan^{-1}(\frac{v_y}{v_x})$

$\theta = tan^{-1}(\frac{-12.8}{-9.89})$

$\theta = -127.7 degree$

7 0

3 years ago

The 26-kg sphere c is released from rest when θ = 0∘ and the tension in the spring is f = 100 n

aleksandr82 [10.1K]

You are given the mass of a sphere that is 26 kg sphere and it is released from rest when θ = 0°. You are also given the force of the spring that is F = 100 N. You are asked to find the tension of the spring. Imagine that the sphere is connected to a spring. The spring exerts a tension and the spring exerts gravitational pull. This will follow the second law of newton.

T - F = ma
T = ma + F
T = 26kg (9.81m/s²) + 100 N
T = 355.06 N

5 0

4 years ago

The two main states of mechanical energy are ___________ and potential energy.

shutvik [7]

<span>Potential energy and Kinetic energy</span>

6 0

3 years ago

Read 2 more answers

A 52.5-turn circular coil of radius 5.35 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0

krek1111 [17]

Answer:

5.43 x 10^-3 Nm

Explanation:

N = 52.5, radius, r = 5.35 cm = 0.0535 m, B = 0.455 T, I = 25.3 mA = 0.0253 A

Torque = N I A B Sin theta

Here, theta = 90 degree

Torque = 52.5 x 0.0253 x 3.14 x 0.0535 x 0.0535 x 0.455

Torque = 5.43 x 10^-3 Nm

6 0

3 years ago