Look at the distance-time graph for a remote-controlled car. What is the speed of the car?

blagie [28]

Place the next vector with its tail at the previous vector's head. ... To subtract vectors, proceed as if adding the two vectors, but flip the vector to be subtracted across the axes and then join it tail to head as if adding. Adding or subtracting any number of vectors yields a resultant vector.

Explanation:

8 0

2 years ago

You want to go jogging at a nearby park. your mother says a thunderstorm is five miles away. the safest thing to do in this situ

sdas [7]

Answer:

"find an indoor place to exercise"

I think that golf courses call in the golfers if there is a lightning strike within 5 miles of the golf course sensor.

7 0

1 year ago

A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.76 m/s2 for t1 = 20 s. At that point the

alex41 [277]

Answer:

(a) $v_1 = a_1t_1 = 1.76 t_1$

(b) It won't hit

(c) 110 m

Explanation:

(a) the car velocity is the initial velocity (at rest so 0) plus product of acceleration and time t1

$v_1 = v_0 + a_1t_1 = 0 +1.76t_1 = 1.76t_1$

(b) The velocity of the car before the driver begins braking is

$v_1 = 1.76*20 = 35.2m/s$

The driver brakes hard and come to rest for t2 = 5s. This means the deceleration of the driver during braking process is

$a_2 = \frac{\Delta v_2}{\Delta t_2} = \frac{v_2 - v_1}{t_2} = \frac{0 - 35.2}{5} = -7.04 m/s^2$

We can use the following equation of motion to calculate how far the car has travel since braking to stop

$s_2 = v_1t_2 + a_2t_2^2/2$

$s_2 = 35.2*5 - 7.04*5^2/2 = 88 m$

Also the distance from start to where the driver starts braking is

$s_1 = a_1t_1^2/2 = 1.76*20^2/2 = 352$

So the total distance from rest to stop is 352 + 88 = 440 m < 550 m so the car won't hit the limb

(c) The distance from the limb to where the car stops is 550 - 440 = 110 m

8 0

2 years ago

A student evaluates a weight loss program by calculating the number of times he would need to climb a 14.0 m high flight of step

Liula [17]

Answer:

400 trips

Explanation:

Mechanical energy needed to climb 14 m by a man of 68 kg

= mgh

= 68 x 9.8 x 14

= 9330 J

1 Kg of fat releases 3.77 x 10⁷ J of energy

.45 kg of fat releases 1.6965 x 10⁷ J of energy

22% is converted into mechanical energy

so 22% of 1.6965 x 10⁷ J

= 3732.3 x 10³ J of mechanical energy will be available for mechanical work.

one trip of climbing of 14 m requires 9330 J of mechanical energy

no of such trip possible with given mechanical energy

= 3732.3 x 10³ / 9330

= 400 trips

7 0

3 years ago

A circular loop of wire with a diameter of 0.626 m is rotated in a uniform electric field to a position where the electric flux

Mila [183]

Answer:

C) 2.44 × 106 N/C

Explanation:

The electric flux through a circular loop of wire is given by

$\Phi = EA cos \theta$

where

E is the electric field

A is the cross-sectional area

$\theta$ is the angle between the direction of the electric field and the normal to A

The flux is maximum when $\theta=0^{\circ}$ , so we are in this situation and therefore $cos \theta =1$ , so we can write

$\Phi = EA$

Here we have:

$\Phi = 7.50\cdot 10^5 N/m^2 C$ is the flux

d = 0.626 m is the diameter of the coil, so the radius is

r = 0.313 m

and so the area is

$A=\pi r^2 = \pi (0.313 m)^2=0.308 m^2$

And so, we can find the magnitude of the electric field:

$E=\frac{\Phi}{A}=\frac{7.50\cdot 10^5 Nm^2/C}{0.308 m^2}=2.44\cdot 10^6 N/C$

3 0

3 years ago