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3 years ago

A trunk is dragged 3.0 meters across an attic floor by a force of 2.0N how much positive work is done on the trunk

Physics

1 answer:

Novay_Z [31]3 years ago

8 0

The formula for work is

F*d

Therefore work=2.0N*3.0=6N*m

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Your friends sit in a sled in the snow. If you apply a force of 95 N to them, they have an acceleration of 0.8 m/s2. What is the

ss7ja [257]

F=ma
m=F/a=95/0.8= 118.75kg
your friend is pretty heavy XD

5 0

3 years ago

Read 2 more answers

Describe what a hydrogen bond is. 6th grade answer

r-ruslan [8.4K]

Answer:

<em>Hydrogen bond is the attractive force between the hydrogen attached electronegative atom </em>

Explanation:

8 0

3 years ago

3.1

Ilia_Sergeevich [38]

Answer:

The value is $C = 30729\ c$

Explanation:

From the question we are told that

The power rating of the stove is $P = 4.4 KW = 4.4 *10^{3} \ W$

The duration of its use everyday is $t_1 = 70 \ minutes = 1.167 \ hours$

The rating of the light bulbs is $P_2' = 150 W$

The number is $n = 7$

The power rating of the total bulb is $P_2 = 7 * 150 = 1050 \ W$

The duration of its use everyday is $t_2 = 7 hours$

The power rating of miscellaneous appliance $P_3 = 1.8 \ KW = 1.8 *10^{3} \ W$

The duration of its use everyday is $t_3 = 1 hour$

The power rating of hot water $P = 4 KW = 4 *10^{3} \ W$

The duration of its use everyday is $t_4 = 120 \ minutes = 2 hours$

Generally the total electrical energy used in 1 month is mathematically represented as

$E = P_1 * t_1 * 30 + P_2 * t_2 * 30 + P_3 * t_3 * 30+ P_4 * t_4 * 30$

=> $E = 4.4*10^{3} * 1.167 * 30 + 1050 * 7 * 30 + 1.8 *1 * 30+ 4*10^{3} * 2 * 30$

=> $E = 614598 \ W \cdot h$

=> $E = 614.6 \ K W \cdot h$

Generally the monthly electricity bill is mathematically represented as

$C = 614.6 * 50$

=> $C = 30729\ c$

8 0

2 years ago

Par 1/2

BaLLatris [955]

part 1

mass = ρ x V

mass = 1739 kg/m³ x 3.8 km³ = 6608.2 kg

PE (potential energy)= mgh

PE = 6608.2 kg x 9.81 x 403

PE = 2.61 x 10⁷ J

part 2

megaton of TNT (Mt) =4.2 x 10¹⁵ J

convert PE to Mt:

2.61 x 10⁷ J : 4.2 x 10¹⁵ J = 6.21 x 10⁻⁹ Mt

4 0

3 years ago

A uranium ion and an iron ion are separated by a distance of =61.10 nm. The uranium atom is singly ionized; the iron atom is dou

scoray [572]

Answer:

Explanation:

Charge on uranium ion = charge of a single electron

= 1.6 x 10⁻¹⁹ C

charge on doubly ionised iron atom = charge of 2 electron

= 2 x 1.6 x 10⁻¹⁹ C = 3.2 x 10⁻¹⁹ C

Let the required distance from uranium ion be d .

force on electron at distance d from uranium ion

= 9 x 10⁹ x 1.6 x 10⁻¹⁹ / r²

force on electron at distance 61.10 x 10⁻⁹ - r from iron ion

= 9 x 10⁹ x 3.2 x 10⁻¹⁹ / (61.10 x 10⁻⁹ - r )²

For equilibrium ,

9 x 10⁹ x 1.6 x 10⁻¹⁹ / r² = 9 x 10⁹ x 3.2 x 10⁻¹⁹ / (61.10 x 10⁻⁹ - r )²

2 d² = (61.10 x 10⁻⁹ - r )²

1.414 r = 61.10 x 10⁻⁹ - r

2.414 r = 61.10 x 10⁻⁹

r = 25.31 nm .

4 0

3 years ago