An alpha particle is identical to a:

Two forces Upper FSubscript Upper A Baseline Overscript right-arrow EndScripts and Upper FSubscript Upper B Baseline Overscript

Pavel [41]

Answer:

Part a)

$F_A = 4.59 N$

Part B)

$F_B = 1.28 N$

Explanation:

As we know that when both the forces are acting on the object in same direction then we will have

$F_A + F_B = ma$

as we know that

$a = 0.554 m/s^2$

m = 10.6 kg

now we will have

$F_A + F_B = 10.6(0.554)$

$F_A + F_B = 5.87 N$

Now two forces are in opposite direction then we have

$F_A - F_B = 10.6(0.313)$

$F_A - F_B = 3.32 N$

Part A)

Now we will have from above two equation

$F_A = 4.59 N$

Part B)

Similarly for other force we have

$F_B = 1.28 N$

5 0

3 years ago

To practice Problem-Solving Strategy 7.2 Problems Using Mechanical Energy II. The Great Sandini is a 60.0-kg circus performer wh

sp2606 [1]

Answer:

$v = 15.45 m/s$

Explanation:

As per mechanical energy conservation we can say that here since friction is present in the barrel so we will have

Work done by friction force = Loss in mechanical energy

so we will have

$W_f = (U_i + K_i) - (U_f + K_f)$

here we know that

$W_f = F_f . d$

$W_f = 40 \times 4$

$W_f = 160 J$

Initial compression in the spring is given as

$F = kx$

$4400 = 1100 x$

$x = 4 m$

now from above equation

$W_f = (\frac{1}{2}kx^2 + 0) - (mgh + \frac{1}{2}mv^2)$

$160 = (\frac{1}{2}1100(4^2) + 0) - (60 \times 9.8\times 2.50 + \frac{1}{2}(60)v^2)$

$160 = 8800 - 1470 - 30 v^2$

$v = 15.45 m/s$

3 0

3 years ago

The current theory of the structure of the Earth, called plate tectonics, tells us that the continents are in constant motion.

lesya [120]

Answer:

(a) m = 1.6 x 10²¹ kg

(b) K.E = 2.536 x 10¹¹ J

(c) v = 7.12 x 10⁵ m/s

Explanation:

(a)

First we find the volume of the continent:

V = L*W*H

where,

V = Volume of Slab = ?

L = Length of Slab = 4450 km = 4.45 x 10⁶ m

W = Width of Slab = 4450 km = 4.45 x 10⁶ m

H = Height of Slab = 31 km = 3.1 x 10⁴ m

Therefore,

V = (4.45 x 10⁶ m)(4.45 x 10⁶ m)(3.1 x 10⁴ m)

V = 6.138 x 10¹⁷ m³

Now, we find the mass:

m = density*V

m = (2620 kg/m³)(6.138 x 10¹⁷ m³)

m = 1.6 x 10²¹ kg

(b)

The kinetic energy will be:

K.E = (1/2)mv²

where,

v = speed = (1 cm/year)(0.01 m/1 cm)(1 year/365 days)(1 day/24 h)(1 h/3600 s)

v = 3.17 x 10⁻¹⁰ m/s

Therefore,

K.E = (1/2)(1.6 x 10²¹ kg)(3.17 x 10⁻¹⁰ m/s)²

K.E = 2.536 x 10¹¹ J

(c)

For the same kinetic energy but mass = 77 kg:

K.E = (1/2)mv²

2.536 x 10¹¹ J = (1/2)(77 kg)v²

v = √(2)(2.536 x 10¹¹ J)

v = 7.12 x 10⁵ m/s

7 0

3 years ago

Two identical charges q placed 2.0 mapart exert forces of magnitude 4.0 N on each other What is the value of the charge q? a) q

katen-ka-za [31]

Answer:

c) $4.2*10^{-5}C$

Explanation:

Coulomb's law says that the force exerted between two charges is inversely proportional to the square of distance between them, and is given by the expression:

$F=\frac{kq_{1}q_{2}}{r^{2}}$

where k is a proportionality constant with the value $k=9*10^{9}\frac{Nm^{2}}{C^{2}}$

In this case $q_{1}=q_{2}=q$ , so we have:

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

Solving the equation for q, we have:

$kq^{2}=Fr^{2}$

$q^{2}=\frac{Fr^{2}}{k}$

$q=\sqrt{\frac{Fr^{2}}{k}}$

Replacing the given values:

$q=\sqrt{\frac{4.0N*(2.0m)^{2}}{9*10^{-9}\frac{Nm^{2}}{C^{2}}}}$

$q=4.2*10^{-5}C$

3 0

3 years ago

shusha [124]

Answer: The unit of impulse is applied to an object produces an equivalent vector change in its linear momentum, also in the same direction.

Explanation:

4 0

3 years ago