A car travels 60 miles in 3 hours. Calculate the speed.

Barney walks at a velocity of 1.7 meters/second on an inclined plane, which has an angle of 18.5° with the ground. What is the h

Viktor [21]

The velocities and the speed build a triangle, where the 1.7 m/s are the hypotenuse and the x-velocity and y-velocity are the other sides.

So the x-velocity is: speed*cos(angle) 

now plug in 
x=1.7 m/s * cos(18.5)=1.597 m/s

3 0

3 years ago

A transformer has two sets of coils, the primary with N1 = 160 turns and the secondary with N2 = 1400 turns. The input rms volta

vovikov84 [41]

To solve the problem it is necessary to apply the concepts related to the voltage in a coil, through the percentage relationship that exists between the voltage and the number of turns it has.

So things our data are given by

$N_1 = 160$

$N_2 = 1400$

$\Delta V_{1rms} = 62V$

PART A) Since it is a system in equilibrium the relationship between the two transformers would be given by

$\frac{N_1}{N_2} = \frac{\Delta V_{1rms}}{\Delta V_{2rms}}$

So the voltage for transformer 2 would be given by,

$\Delta V_{2rms} = \frac{N_2}{N_1} \Delta V_{1rms}$

PART B) To express the number value we proceed to replace with the previously given values, that is to say

$\Delta V_{2rms} = \frac{N_1}{N_2} \Delta V_{1rms}$

$\Delta V_{2rms} = \frac{1400}{160} 62V$

$\Delta V_{2rms} = 1446.66V$

7 0

3 years ago

What is the resultant force acting on an object if both forces act in the

Flura [38]

Explanation:

Touch and hold a clip to pin it. Unpinned clips will be deleted after 1 hour.Touch and hold a clip to pin it. Unpinned clips will be deleted after 1 hour.

4 0

3 years ago

An irregular shape object has a mass of 19 oz. A graduated cylinder with and initial volume of 33.9 mL. After the object was dro

madreJ [45]

Explanation:

m = 19 oz × (28.3 g/1 oz) = 537.7 g

V = 92.8 mL

$\rho = \dfrac{m}{V}= \dfrac{537.7\:g}{92.8\:mL} = 5.79\:\frac{g}{mL}$

7 0

3 years ago

A spherical bowling ball with mass m = 3.6 kg and radius R = 0.101 m is thrown down the lane with an initial speed of v = 8.7 m/

luda_lava [24]

Answer:

1) The magnitude of the angular acceleration = 67.92 rad/ $s^{2}$

2) Magnitude of the linear acceleration = 2.744 m/ $s^{2}$

3) How long does it take the bowling ball to begin rolling without slipping = 0.906 s

4) How long does it take the bowling ball to begin rolling without slipping = 6.75 m

5) the final velocity is 6.21 m/s

Explanation:

the given information :

Bowling mass m = 3.6 kg

Radius = 0.101 m

Initial speed $v_{0}$ = 8.7 m/s

Coefficient of kinetic friction μ = 0.28

1) he magnitude of the angular acceleration of the bowling ball is

F = m a

$F_{g}$ = μ N , N = m g

$F_{g}$ = μ m g

1) The magnitude of the angular acceleration of the bowling ball as it slides down the lane:

momen inersia of Bowling ball I = (2/5) m $R^{2}$

torque τ = I α

τ = F R

I α = F R

(2/5) m $R^{2}$ α = μ m g R

α = (5 μ g / 2R) μ g R

= (5 x 0.28 x 9.8/ 2 x 0.101)

= 67.92 rad/ $s^{2}$

2) Magnitude of the linear acceleration of the bowling ball as it slides down the lane

F = - $F_{g}$ , $F_{k}$ is the force of kinetic friction

m a = - μ m g, remove m

the magnitude of linear accelaration is

a = μ g

= (0.28) (9.8)

= 2.744 m/ $s^{2}$

3) The bowling ball takes time to begin rolling without slipping:

The linear speed, $v_{t}$ = $v_{0}$ - a t

$v_{t}$ = $v_{0}$ - μ g t

the angular speed, ω = ω0 + α t

ω = ω0 + (5 μ g/2R ) t

$v_{t}$ = ω R

$v_{0}$ - μ g t = ω0 R + (5 μ g/2R ) t R

7 μ g t/2 = $v_{0}$ + ω0 R

hence,

t = (2 $v_{0}$ + ω0 R)/ 7 μ g

ω0 = 0 (no initial spin), therefore

t = 2 $v_{0}$ / 7 μ g

= 2 x 8.7 / 7 (0.28) (9.8)

= 0.906 s

4) How long it takes for the bowling ball to begin rolling without slipping, S

S = $v_{0}$ t - (1/2) a $t^{2}$

= (8.7) (0.906) - (1/2) (2.744) $0.906^{2}$

= 6.75 m

5) The final velocity

$v_{t}$ = $v_{0}$ - a t

$v_{t}$ = 8.7 - (2.744) (0.906)

$v_{t}$ = 6.21 m/s

4 0

3 years ago