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3 years ago

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A car of mass 1230 kg is on an icy driveway inclined at an angle of 39°. The acceleration of gravity is 9.8 m/s². θ If the incli

ne is friction-less, what is the acceleration of the car? Answer in units of m/s².

1 answer:

tamaranim1 [39]3 years ago

6 0

Answer:

Acceleration of the car, $a=6.16\ m/s^2$

Explanation:

It is given that,

Mass of the car, m = 1230 kg

The car is inclined at an angle of 39 degrees.

The acceleration of gravity acting on the car is $9.8\ m/s^2$ . The component of car parallel to inclined planes is :

$W_x=mg\ sin\theta$

Net force acting on it is :

$ma=mg\ sin\theta$

$a=g\ sin\theta$

$a=9.8\times \ sin(39)$

$a=6.16\ m/s^2$

So, the acceleration of the car is $6.16\ m/s^2$ . Hence, this is the required solution.

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A rubber ball and a lump of clay have equal mass. They are thrown with equal speed against a wall. The ball bounces back with ne

gregori [183]

Answer:

The ball experiences the greater momentum change

Explanation:

The momentum change of each object is given by:

$\Delta p = m \Delta v= m (v-u)$

where

m is the mass of the object

v is the final velocity

u is the initial velocity

Both objects have same mass m and same initial velocity u. So we have:

- For the ball, the final velocity is

$v=-u$

Since it bounces back (so, opposite direction --> negative sign) with same speed (so, the magnitude of the final velocity is still u). So the change in momentum is

$\Delta p=m(v-u)=m((-u)-u)=-2mu$

- For the clay, the final velocity is

$v=0$

since it sticks to the wall. So, the change in momentum is

$\Delta p = m(v-u)=m(0-u)=-mu$

So we see that the greater momentum change (in magnitude) is experienced by the ball.

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3 years ago

Please help<br> Right answers please<br> Will mark brainliest

balandron [24]

Answer:

a. 45 N. / b. 0.08 m/s^2. / c. 102 N

F = ma

F = 15(3)

F = 45 newtons

F/m = a

20/250 = a

0.08 m/s^2 = a

R = ma

R =1.5(68)

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3 years ago

Assume: The bullet penetrates into the block and stops due to its friction with the block. The compound system of the block plus

tino4ka555 [31]

Answer:

1)4.7334J

2)225.4m/s

Explanation:

v= the Velocity of both the bullet and the block after collision=?

H= Height of the bullet along circular arc= 10cm=0.1m

g= acceleration due to gravity= 9.81m/s^2

R= Radius of the circular arc= 18cm= 0.18m

m= Mass of the bullet= 30g= 0.03kg

M= Mass of the block = 4.8 kg

Using the law of conservation of energy

Potential energy of the system= Kinectic energy of the system

1/2 mv^2= mgh..............eqn(1)

But we have two mass m and M

We can write eqn(1) as

0.5(m+M)v^2= (m+M)gh ...........eqn(2)

If we make "v" subject of the formula we have

v = √2gh

Then substitute the values we have

= √2 x 9.81 x 0.1 = 1.40m/s

1) We can now calculate the total energy of the system after collision as

KE = 1/2(m+M)v^2

= 1/2 x (0.03+4.8) x (1.40)^2

KE = 4.7334J

Hence, the total energy of the composite system at any time after the collision is 4.7334J

2)to determine the initial velocity of the bullet.

From law of momentum conservation, which can be expressed as

m1u1+m2u2=(m1+m2)v

Where the initial Velocity of the bullet u1= ?

Final velocity of the bullet = 0

the Velocity of both the bullet and the block after collision=v= 1.40m/s

(0.03×u1) +(u×0)= (4.8+0.03)1.4

0.03u1=6.762

U1=225.4m/s

Hence, the initial velocity of the bullet is 225.4m/s

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SashulF [63]

Answer:

Part a)

$F = 0$

Part b)

$F = 0.25 N$

Part c)

$F = 0.25 N$

Part d)

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$F_{net} = 0$

Explanation:

As we know that force on a current carrying wire is given as

$\vec F = i(\vec L \times \vec B)$

now we have

Part a)

current in side 166 cm and magnetic field is parallel

so we have

$F = i(\vec L \times \vec B)$

here we know that L and B is parallel to each other so

$F = 0$

Part b)

For 68.1 cm length wire we have

$F = iLB sin\theta$

here we know that

$cos\theta = \frac{68.1}{166}$

$\theta = 65.8$

so we have

$F = (4.03)(0.681)(99.3 \times 10^{-3})sin65.8$

$F = 0.25 N$

Part c)

For 151 cm length wire we have

$F = iLB sin\phi$

here we know that

$cos\phi = \frac{151}{166}$

$\theta = 24.5$

so we have

$F = (4.03)(1.51)(99.3 \times 10^{-3})sin24.5$

$F = 0.25 N$

Part d)

Net force on a closed loop in uniform magnetic field is always ZERO

$F_{net} = 0$

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BabaBlast [244]

Sowers has spent her career fighting for gender equality and it passionate about it .

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