Ask question

Ask question

All categories

2 years ago

5

A dog on a skateboard applies a force of 84 N. He accelerates at 3.7 m/s2 and there is a friction force of 33 N. What is the dog

’s mass?

1 answer:

Delicious77 [7]2 years ago

8 0

Answer:

34.84727272727273 pounds

Explanation:

84 * 3.7m squared = 11.94.96 / 33 = 34.84727272727273

You might be interested in

An electron is accelerated through 2400 V from rest and then enters a region in which there is a uniform 1.70 T magnetic field.

aalyn [17]

Answer:

Explanation:

Let v be the velocity acquired by electron in electric field

V q = 1/2 m v²

V is potential difference applied on charge q , m is mass of charge , v is velocity acquired

2400 x 1.6 x 10⁻¹⁹ = .5 x 9.1 x 10⁻³¹ x v²

v² = 844 x 10¹²

v = 29.05 x 10⁶ m /s

Maximum force will be exerted on moving electron when it moves perpendicular to magnetic field .

Maximum force = Bqv , where B is magnetic field , q is charge on electron and v is velocity of electron

= 1.7 x 1.6 x 10⁻¹⁹ x 29.05 x 10⁶

= 79.02 x 10⁻¹³ N .

Minimum force will be zero when electron moves along the direction of magnetic field .

5 0

2 years ago

So... if bedbugs live in beds, where do cockroaches live?

Deffense [45]

Answer:

oo.p i wish I could answer that

Explanation:

6 0

2 years ago

Read 2 more answers

1. A sample consisting of 1.0 mol CaCO3 (s) was heated to 800oC when it is decomposed. The heating was carried out in a containe

ExtremeBDS [4]

Expansion work against constant external pressure: w=-pex Δ Δ V 3. The attempt at a solution . I tried following that. Because Vf>>Vi, and Vf=nRT/pex, then w=-pex x nRT/pex=-nRT (im assuming n is number of moles of CO2?). 1 mole of CaCO3 makes 1 mole of CO2, so plugging in numbers, I get 8.9kJ, although I dont use the 1 atm pressure at all

6 0

2 years ago

The half-life of a certain radioactive substance is 12 hours. There are 19 grams present initially. a. Express the amount of su

neonofarm [45]

Answer:

(a) $N=19\times e^{-\lambda t}$

(b) 15 hours

Explanation:

half life, T = 12 hours

No = 19 g

(a) Let N be the amount remaining after time t.

Let λ be the decay constant.

$\lambda =\frac {0.6931}{T}$

The equation of radioactivity used here is given by

$N=N_{o}e^{-\lambda t}$

$N=19\times e^{-\lambda t}$

(b) N = 8 gram

Substitute the values in above equation

$\lambda =\frac {0.6931}{12}$

λ = 0.0577 per hour

So, $8=19\times e^{-0.577t}$

$e^{-0.0577t}=0.421$

Take natural log on both the sides

- 0.0577 t = - 0.865

t = 15 hours

4 0

2 years ago

A train, traveling at a constant speed of 22.0 m/s, comes to an incline with a constant slope. While going up the incline, the t

Charra [1.4K]

Answer:

123.30 m

Explanation:

Given

Speed, u = 22 m/s

acceleration, a = 1.40 m/s²

time, t = 7.30 s

From equation of motion,

v = u + at

where,

v is the final velocity

u is the initial velocity

a is the acceleration

t is time

V = at + U

using equation v - u = at to get line equation for the graph of the motion of the train on the incline plane

$V_{x} = mt + V_{o}$ where m is the slope

Comparing equation (1) and (2)

$V = V_{x}$

a = m

$U = V_{o}$

Since the train slows down with a constant acceleration of magnitude 1.40 m/s² when going up the incline plane. This implies the train is decelerating. Therefore, the train is experiencing negative acceleration.

a = - 1.40 m/s²

Sunstituting a = - 1.40 m/s² and u = 22 m/s

$V_{x} = -1.40t + 22$

$V_{x} = -1.40(7.30) + 22$

$V_{x} = -10.22 + 22$

$V_{x} = 11. 78 m/s$

The speed of the train at 7.30 s is 11.78 m/s.

The distance traveled after 7.30 sec on the incline is the area cover on the incline under the specific interval.

Area of triangle + Area of rectangle

$[\frac{1}{2} * (22 - 11.78) * (7.30)] + [(11.78 - 0) * (7.30)]$

= 37.303 + 85.994

= 123. 297 m

≈ 123. 30 m

4 0

2 years ago

Read 2 more answers