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3 years ago

5

What would be the path of an object thrown in the air if there was no gravity?

2 answers:

alex41 [277]3 years ago

8 0

Answer:D) in a straight line.

Hey

Newton's first law says that if an object is at rest it will stay at rest. But if it is moving it will continue moving in a straight line if there is no external force. If there where no gravity the object that you throw will keep going in a straight line according to Newton's first law.

Svetllana [295]3 years ago

7 0

Answer:

What would be the path of an object thrown in the air if there was no gravity?

the answer is D

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Find the magnitude and direction of an electric field that exerts a 4.80×10−17N westward force on an electron. (b) What magnitud

jok3333 [9.3K]

Explanation:

(a) E = F/q

E = 4.8×10^-17/1.6×10^-19

E = 300 N/C

(b) same magnitude of electric field is exerted on proton

4 0

3 years ago

A 20 cm-radius ball is uniformly charged to 71 nC.

artcher [175]

Answer:

Part a)

$\rho = 2.12\mu C/m^3$

Part b)

$q_1 = 1.11 nC$

$q_2 = 8.88 nC$

$q_3 = 71 nC$

Part c)

$E_1 = 3996 N/C$

$E_2 = 7992 N/C$

$E_3 = 15975 N/C$

Explanation:

Part a)

As we know that charge density is the ratio of total charge and total volume

So here the volume of the charge ball is given as

$V = \frac{4}{3}\pi R^3$

$V = \frac{4}{3}\pi(0.20)^3$

$V = 0.0335 m^3$

now the charge density of the ball is given as

$\rho = \frac{71 nC}{0.0335} = 2.12\mu C/m^3$

Part b)

Now the charge enclosed by the surface is given as

$q = \rho V$

at radius of 5 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.05)^3$

$q = 1.11 nC$

at radius of 10 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.10)^3$

$q = 8.88 nC$

at radius of 20 cm

$q = 71 nC$

Part c)

As we know that electric field is given as

$E = \frac{kq}{r^2}$

so we have electric field at r = 5 cm

$E_1 = \frac{(9\times 10^9)(1.11 nC)}{0.05^2}$

$E_1 = 3996 N/C$

electric field at r = 10 cm

$E_2 = \frac{(9\times 10^9)(8.88 nC)}{0.10^2}$

$E_2 = 7992 N/C$

electric field at r = 20 cm

$E_3 = \frac{(9\times 10^9)(71 nC)}{0.20^2}$

$E_3 = 15975 N/C$

3 0

3 years ago

A truck travels 1430 m uphill along a road that makes a constant angle of 5.76◦ with the horizontal.

Alex73 [517]

Answer:

The horizontal component of displacement is d' = 1422.7 m

Explanation:

Given data,

The distance covered by the truck, d = 1430 m

The angle formed with the horizontal, Ф = 5.76°

The displacement is a vector quantity.

The horizontal component of displacement is given by,

d' = d cos Ф

= 1430 cos 5.76°

= 1422.7 m

Hence, the horizontal component of displacement is d' = 1422.7 m

5 0

3 years ago

A jetliner flies at a constant speed covering 467 miles in 3.3 hours. What is the speed of the plane in miles per hour?

stellarik [79]

Answer:

141.152 miles per hour is the speed of the plane in miles per hour

Explanation:

Speed of plane = Total distance travelled/total time taken -

v = D/t

Substituting the given values in the above equation, we get

v = 467/3.3 miles /hour

v = 141.152 miles per hour

141.152 miles per hour is the speed of the plane in miles per hour

8 0

3 years ago

A block-and-tackle pulley hoist is suspended in a warehouse by ropes of lengths 2 m and 3 m. the hoist weighs 430 n. the ropes,

ICE Princess25 [194]

Refer to the diagram shown below.

For horizontal equilibrium,
T₃ cos38 = T₂ cos 50
0.788 T₃ = 0.6428 T₂
T₃ = 0.8157 T₂ (1)

For vertical equilibrium,
T₂ sin 50 + T₃ sin 38 = 430
0.766 T₂ + 0.6157 T₃ = 430
1.2441 T₂ + T₃ = 698.392 (2)

Substitute (1) into (2).
(1.2441 + 0.8157) T₂ = 698.392
T₂ = 339.058 N
T₃ = 0.8157(399.058) = 276.571 N

Answer:
T₂ = 339.06 N
T₃ = 276.57 N

7 0

3 years ago