Which type of cloud is often taken to be a sign of fair weather?.

A stone is thrown straight up. when it reaches it’s highest point, _____

Sati [7]

both it velocity and acceleration is zero.

8 0

2 years ago

James accelerates his skate board uniformly along a straight road from rest to 10 m/s in 4 seconds. What is James Acceleration?

lawyer [7]

Given:

u(initial velocity)=0

v(final velocity)= 10 m/s

t= 4 sec

Now we know that

v= u + at

Where v is the final velocity

u is the initial velocity

a is the acceleration measured in m/s^2

t is the time measured in sec

10=0+ax4

a=10/4

a=2.5 m/s^2

4 0

3 years ago

A blow-dryer and a vacuum cleaner each operate with a voltage of 120 V. The current rating of the blow-dryer is 12 A, while that

andreev551 [17]

Answer:

a) 1450watts

b) 564watts

c) 1.11

Explanation:

Power consumed = IV

I is the current rating

V is the operating voltage

If a blow-dryer and a vacuum cleaner each operate with a voltage of 120 V and the current rating of the blow-dryer is 12 A, while that of the vacuum cleaner is 4.7 A then their individual power rating is calculated thus;

a) For blow-dryer

Operating voltage = 120V

Its current rating = 12A

Power consumed = IV

= 120×12

= 1440watts

b) For vacuum cleaner:

Operating voltage is the same as that of blow dryer = 120V

Its current rating = 4.7A

Power consumed = IV

= 120×4.7

= 564watts

c) Energy used = Power consumed × time taken

Energy used = Power × time

Energy used by blow dryer = 1440×20×60

= 1,728,000Joules

Energy used up by vacuum cleaner = 564×46×60

= 564×2760

= 1,556,640Joules

Ratio of the energy used by the blow-dryer in 20 minutes to the energy used by the vacuum cleaner in 46 minutes will be 1,728,000/1,556,640 = 1.11

4 0

3 years ago

A gardener mows a lawn with an old-fashioned push mower. The handle of the mower makes an angle of 320 with the surface of the l

Virty [35]

Answer:

N = 337.96 N

Explanation:

∅ = 32º

F = 249 N

m = 21 Kg

N = ?

We can apply:

∑ F = 0 (↑)

- Fy - W + N = 0 ⇒ N = Fy + W

⇒ F*Sin ∅ + m*g = N

⇒ N = (249 N*Sin32º) + (21 Kg*9.81 m/s²)

⇒ N = 337.96 N (↑)

8 0

3 years ago

A long, rigid conductor, lying along the x-axis, carries a current of 7.0 A in the negative direction. A magnetic field B is pre

Alisiya [41]

Answer:

0.546 $\hat k$

Explanation:

From the given information:

The force on a given current-carrying conductor is:

$F = I ( \L \limits ^ {\to } \times B ^{\to})\\ \\ dF = I(dL\limits ^ {\to } \times B ^{\to})$

where the length usually in negative (x) direction can be computed as

$\L ^ {\to } = -x\hat i \\dL\limits ^ {\to }- dx\hat i$

Now, taking the integral of the force between x = 1.0 m and x = 3.0 m to get the value of the force, we have:

$\int dF = \int ^3_1 I ( dL^{\to} \times B ^{\to})$

$F = I \int^3_1 ( -dx \hat i ) \times ( 4.0 \hat i + 9.0 \ x^2 \hat j)$

$F = I \int^3_1 - 9.0x^2 \ dx \hat k$

$F = I (9.0) \bigg [\dfrac{x^3}{3} \bigg ] ^3_1 \hat k$

$F = I (9.0) \bigg [\dfrac{3^3}{3} - \dfrac{1^3}{3} \bigg ] \hat k$

where;

current I = 7.0 A

$F = (7.0 \ A) (9.0) \bigg [\dfrac{27}{3} - \dfrac{1}{3} \bigg ] \hat k$

$F = (7.0 \ A) (9.0) \bigg [\dfrac{26}{3} \bigg ] \hat k$

F = 546 × 10⁻³ T/mT $\hat k$

F = 0.546 $\hat k$

4 0

3 years ago